题解

121 条题解

  • 4
    @ 2017-09-20 22:39:44

    各位大佬都是深搜啊。。。
    这里补一个广搜的代码
    写丑了不要介意

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std;
    char map[510][510];
    int yan[510][510],dx[300010],dy[300010],bx[5]={0,1,-1,0,0},by[5]={0,0,0,1,-1};
    int main()
    {
        int nx,ny,x,y,i,k,sum=0,nowx,nowy,head=1,tail=1,c;
        cin>>nx>>ny;
        for(i=1;i<=nx;i++)
          for(k=1;k<=ny;k++)
           cin>>map[i][k];
        for(k=1;k<=nx;k++)
         {
            if(map[k][1]=='0')
            yan[k][1]=1;
            else
            continue;
            tail=1,head=1;
            dx[1]=k,dy[1]=1;    
           while(head>=tail)
          {
            nowx=dx[tail];
            nowy=dy[tail];
            for(i=1;i<=4;i++)
            {
                x=nowx+bx[i];
                y=nowy+by[i];
                if(x>0&&x<=nx&&y>0&&y<=ny&&map[x][y]=='0'&&yan[x][y]==0)
                {
                    head++;
                    dx[head]=x;
                    dy[head]=y;
                    yan[x][y]=1;
                }
            }
            tail++;
         }
         }
         for(k=1;k<=nx;k++)
         {
            if(map[k][ny]=='0')
            yan[k][ny]=1;
            else
            continue;
            tail=1,head=1;
            dx[1]=k,dy[1]=ny;   
           while(head>=tail)
          {
            nowx=dx[tail];
            nowy=dy[tail];
            for(i=1;i<=4;i++)
            {
                x=nowx+bx[i];
                y=nowy+by[i];
                if(x>0&&x<=nx&&y>0&&y<=ny&&map[x][y]=='0'&&yan[x][y]==0)
                {
                    head++;
                    dx[head]=x;
                    dy[head]=y;
                    yan[x][y]=1;
                }
            }
            tail++;
         }
         }
         for(k=1;k<=ny;k++)
         {
            if(map[1][k]=='0')
            yan[1][k]=1;
            else
            continue;
            tail=1,head=1;
            dx[1]=1,dy[1]=k;    
           while(head>=tail)
          {
            nowx=dx[tail];
            nowy=dy[tail];
            for(i=1;i<=4;i++)
            {
                x=nowx+bx[i];
                y=nowy+by[i];
                if(x>0&&x<=nx&&y>0&&y<=ny&&map[x][y]=='0'&&yan[x][y]==0)
                {
                    head++;
                    dx[head]=x;
                    dy[head]=y;
                    yan[x][y]=1;
                }
            }
            tail++;
         }
         }
         for(k=1;k<=ny;k++)
         {
            if(map[nx][k]=='0')
            yan[nx][k]=1;
            else
            continue;
            tail=1,head=1;
            dx[1]=nx,dy[1]=k;   
           while(head>=tail)
          {
            nowx=dx[tail];
            nowy=dy[tail];
            for(i=1;i<=4;i++)
            {
                x=nowx+bx[i];
                y=nowy+by[i];
                if(x>0&&x<=nx&&y>0&&y<=ny&&map[x][y]=='0'&&yan[x][y]==0)
                {
                    head++;
                    dx[head]=x;
                    dy[head]=y;
                    yan[x][y]=1;
                }
            }
            tail++;
         }
         }
         for(i=1;i<=nx;i++)
          for(k=1;k<=ny;k++)
           if(map[i][k]=='0'&&yan[i][k]==0)
             sum++;
         cout<<sum;
         return 0;
    }
    
  • 1
    @ 2019-05-29 19:07:19

    BFS,理解为水从边框的0向里漫,被水淹没的标记为+,最后统计没被淹没的0的个数。

    #include <iostream>
    #include <queue>
    
    using namespace std;
    
    typedef struct
    {
        int x,y;
    }point;
    
    int m,n;
    char c[500][500];
    queue<point> que;
    int ans=0;
    
    int main()
    {
        cin>>m>>n;
        int i,j;
        point it,pu;
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                cin>>c[i][j];
            }
        }
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(i==0||i==m-1||j==0||j==n-1)
                {
                    if(c[i][j]=='0')
                    {
                        it.x=i;
                        it.y=j;
                        que.push(it);
                        while(!que.empty())
                        {
                            it=que.front();
                            que.pop();
                            c[it.x][it.y]='+';
                            if(it.x>0)
                            {
                                if(c[it.x-1][it.y]=='0')
                                {
                                    pu.x=it.x-1;
                                    pu.y=it.y;
                                    que.push(pu);
                                }
                            }
                            if(it.x<m-1)
                            {
                                if(c[it.x+1][it.y]=='0')
                                {
                                    pu.x=it.x+1;
                                    pu.y=it.y;
                                    que.push(pu);
                                }
                            }
                            if(it.y>0)
                            {
                                if(c[it.x][it.y-1]=='0')
                                {
                                    pu.x=it.x;
                                    pu.y=it.y-1;
                                    que.push(pu);
                                }
                            }
                            if(it.y<n-1)
                            {
                                if(c[it.x][it.y+1]=='0')
                                {
                                    pu.x=it.x;
                                    pu.y=it.y+1;
                                    que.push(pu);
                                }
                            }
                        }
                    }
                }
            }
        }
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(c[i][j]=='0')
                {
                    ans++;
                }
            }
        }
        cout<<ans<<endl;
        return 0;
    }
    
  • 1
    @ 2017-11-04 20:12:47

    简单的DFS。◕‿◕。

    #include<iostream>
    using namespace std;
    int x,y,k=0;
    char a[600][600];
    void sub(int,int);
    void sub(int i,int j)
    {
        a[i][j]='s';
        if(a[i-1][j]=='0')
        sub(i-1,j);
        if(a[i+1][j]=='0')
        sub(i+1,j);
        if(a[i][j-1]=='0')
        sub(i,j-1);
        if(a[i][j+1]=='0')
        sub(i,j+1);
    }
    int main()
    {
        cin>>x>>y;
        for(int i=1;i<=x;++i)
        {
            for(int j=1;j<=y;++j)
            {
                cin>>a[i][j];
            }
        } 
        for(int i=1;i<=y;++i)
        if(a[1][i]=='0')
        sub(1,i);
        for(int i=1;i<=y;++i)
        if(a[x][i]=='0')
        sub(x,i);
        for(int i=1;i<=x;++i)
        if(a[i][1]=='0')
        sub(i,1);
        for(int i=1;i<=x;++i)
        if(a[i][y]=='0')
        sub(i,y);
        for(int i=1;i<=x;++i)
        {
            for(int j=1;j<=y;++j)
            {
                if(a[i][j]=='0')
                k++;
            }
        }
        cout<<k;
        return 0;
    }
    
  • 1
    @ 2016-08-10 18:12:05

    /*
    灌水法深搜,注意边界处理。还需要判断是否曾来过,减少不必要搜索。
    */
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <iomanip>
    #include <cstdlib>
    using namespace std;

    int n,m;
    char a[502][502];
    int zx[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

    void dfs(int x,int y)
    {
    if(x>n||x<1||y<1||y>m)
    return;
    if(a[x][y]==1)//此句应在下一句前
    return;
    if(a[x][y]=='0')
    a[x][y]=1;
    if(a[x][y]=='*')
    return;
    for(int i=0;i<4;i++)
    {
    int newx=x+zx[i][0];
    int newy=y+zx[i][1];
    dfs(newx,newy);
    }
    }

    int main()
    {
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    cin>>a[i]+1;
    for(int i=1;i<=n;i++)
    {

    dfs(i,1);

    dfs(i,m);
    }
    for(int i=1;i<=m;i++)
    {

    dfs(1,i);

    dfs(n,i);
    }
    int ans=0;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    if(a[i][j]=='0')
    ans++;
    cout<<ans<<endl;
    return 0;
    }

  • 0
    @ 2017-10-05 02:28:07

    so water

    #include <cstdio>
    #include <cstdlib>
    #include <algorithm>
    #include <iostream>  
    #include <cmath>  
    #include <cstring>  
    #include <set>  
    using namespace std;
    int a[510][510]={0},all=0,temp=0,e=1,m,n;
    int xku[5]={0,1,-1,0,0},yku[5]={0,0,0,1,-1};
    void bfs(int x,int y)
    {
        for(int i=1;i<=4;i++)
        {
            int n_x=x+xku[i],n_y=y+yku[i];
            if(n_x>0&&n_y>0&&n_x<=n&&n_y<=m)
            {
                if(!a[n_x][n_y])
                {
                    temp+=e;
                    a[n_x][n_y]=1;
                    bfs(n_x,n_y);
                }
            }
            else
            {
                e=0;
                temp=0;
            }
        }
    }
    int main()
    {
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            char line[510]={0};
            scanf("%s",line);
            for(int j=1;j<=m;j++)
            {
                if(line[j-1]=='*')
                    a[i][j]=1;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]==0)
                {
                    temp=1;
                    e=1;
                    a[i][j]=1;
                    bfs(i,j);
                    all+=temp;
                }   
            }
        }
        printf("%d",all);
        return 0;  
    }  #include <cstdio>
    #include <cstdlib>
    #include <algorithm>
    #include <iostream>  
    #include <cmath>  
    #include <cstring>  
    #include <set>  
    using namespace std;
    int a[510][510]={0},all=0,temp=0,e=1,m,n;
    int xku[5]={0,1,-1,0,0},yku[5]={0,0,0,1,-1};
    void bfs(int x,int y)
    {
        for(int i=1;i<=4;i++)
        {
            int n_x=x+xku[i],n_y=y+yku[i];
            if(n_x>0&&n_y>0&&n_x<=n&&n_y<=m)
            {
                if(!a[n_x][n_y])
                {
                    temp+=e;
                    a[n_x][n_y]=1;
                    bfs(n_x,n_y);
                }
            }
            else
            {
                e=0;
                temp=0;
            }
        }
    }
    int main()
    {
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            char line[510]={0};
            scanf("%s",line);
            for(int j=1;j<=m;j++)
            {
                if(line[j-1]=='*')
                    a[i][j]=1;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]==0)
                {
                    temp=1;
                    e=1;
                    a[i][j]=1;
                    bfs(i,j);
                    all+=temp;
                }   
            }
        }
        printf("%d",all);
        return 0;  
    }  
    
  • 0
    @ 2016-09-05 21:11:39

    DFS 把边界设为水,vis[]标记
    #include <cstdio>
    #include <cstring>
    #define is_ok(i,j) (0<=i&&i<=n+1&&0<=j&&j<=m+1&&vis[i][j]==0&&A[i][j]!='*')

    int n,m;
    int star=0,count=0;
    char A[600][600];

    int vis[600][600]={0};

    void dfs(int x,int y){
    if(A[x][y]=='0')
    count++;
    vis[x][y]=1;
    if(is_ok(x+1,y)) dfs(x+1,y);
    if(is_ok(x-1,y)) dfs(x-1,y);
    if(is_ok(x,y+1)) dfs(x,y+1);
    if(is_ok(x,y-1)) dfs(x,y-1);
    }

    int main(){
    memset(A,'W',sizeof(A));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++){
    scanf(" %c",&A[i][j]);
    if(A[i][j]=='*')
    star++;
    }
    dfs(0,0);
    printf("%d",m*n-star-count);
    return 0;
    }

  • 0
    @ 2016-08-23 13:08:36

    水题
    var n,m,i,j,ans:longint;
    a:array [0..500,0..500] of char;

    procedure find(x,y:longint);
    begin
    if a[x,y]='0' then
    begin
    a[x,y]:='*';
    find(x-1,y);
    find(x,y-1);
    find(x,y+1);
    find(x+1,y);
    end;
    end;

    begin
    readln(n,m);
    for i:=1 to n do
    begin
    for j:=1 to m do
    read(a[i,j]);
    readln;
    end;
    for i:=1 to n do
    begin
    find(i,1);
    find(i,m);
    end;
    for i:=1 to m do
    begin
    find(1,i);
    find(n,i);
    end;
    for i:=1 to n do
    for j:=1 to m do
    if a[i,j]='0' then
    inc(ans);
    write(ans);
    end.

  • 0
    @ 2015-10-03 17:23:58

    一个方框搜一下全部标记‘*’,看一看剩下数量

  • 0
    @ 2015-08-10 17:37:52

    P1294拯救OIBH总部
    Accepted

    记录信息

    评测状态 Accepted
    题目 P1294 拯救OIBH总部
    递交时间 2015-08-10 17:37:13
    代码语言 C++
    评测机 VijosEx
    消耗时间 141 ms
    消耗内存 1104 KiB
    评测时间 2015-08-10 17:37:14

    评测结果

    编译成功

    foo.cpp: In function 'int main()':
    foo.cpp:33:22: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
    cin >> a[i] + 1;
    ^

    测试数据 #0: Accepted, time = 0 ms, mem = 1076 KiB, score = 10

    测试数据 #1: Accepted, time = 0 ms, mem = 1040 KiB, score = 10

    测试数据 #2: Accepted, time = 15 ms, mem = 1052 KiB, score = 10

    测试数据 #3: Accepted, time = 31 ms, mem = 1104 KiB, score = 10

    测试数据 #4: Accepted, time = 18 ms, mem = 1100 KiB, score = 10

    测试数据 #5: Accepted, time = 31 ms, mem = 1092 KiB, score = 10

    测试数据 #6: Accepted, time = 15 ms, mem = 1060 KiB, score = 10

    测试数据 #7: Accepted, time = 0 ms, mem = 1052 KiB, score = 10

    测试数据 #8: Accepted, time = 31 ms, mem = 1092 KiB, score = 10

    测试数据 #9: Accepted, time = 0 ms, mem = 1064 KiB, score = 10

    Accepted, time = 141 ms, mem = 1104 KiB, score = 100

    代码

    #include <iostream>
    #include <stdio.h>

    using namespace std;

    int n , m;
    char a[500 + 5][500 + 5];
    bool vis[500 + 5][500 + 5];
    int i , j;

    void dfs( int x , int y )
    {
    if( x < 0 || y < 0 || x > n + 1 || y > m + 1 )
    return;
    if( vis[x][y] )
    return;
    if( a[x][y] == '*' )
    return;
    vis[x][y] = 1;
    dfs( x - 1 , y );
    dfs( x + 1 , y );
    dfs( x , y - 1 );
    dfs( x , y + 1 );
    return;
    }

    int ans;

    int main()
    {
    scanf( "%d %d" , &n , &m );
    for( i = 1 ; i <= n ; i++ )
    cin >> a[i] + 1;
    for( i = 1 ; i <= n ; i++ )
    {
    dfs( i , 1 );
    dfs( i , m );
    }
    for( i = 1 ; i <= m ; i++ )
    {
    dfs( 1 , i );
    dfs( n , i );
    }
    for( i = 1 ; i <= n ; i++ )
    for( j = 1 ; j <= m ; j++ )
    if( !vis[i][j] && a[i][j] == '0' )
    ans++;
    cout << ans << endl;
    return 0;
    }
    这坑人的水题

  • 0
    @ 2015-08-02 20:34:33

    program exam1294;
    const dx:array[1..4]of longint=(-1,1,0,0);
    dy:array[1..4]of longint=(0,0,-1,1);
    var n,m,i,j,tot:longint;
    f:array[0..600,0..600]of integer;
    h:array[0..20000,0..2]of longint;
    s:char;

    procedure bfs(p,q:longint);
    var i,t,w,x,y:longint;
    begin
    f[p,q]:=1;
    t:=1;
    w:=1;
    h[1,1]:=p; h[1,2]:=q;
    repeat
    for i:=1 to 4 do
    begin
    x:=h[t,1]+dx[i];
    y:=h[t,2]+dy[i];
    if (x>0)and(x<=n)and(y>0)and(y<=m)and(f[x,y]=0) then
    begin
    inc(w);
    h[w,1]:=x;
    h[w,2]:=y;
    f[x,y]:=1;
    end;
    end;
    inc(t);
    until t>w;
    end;

    begin
    readln(n,m);
    for i:=1 to n do
    begin
    for j:=1 to m do
    begin
    read(s);
    if s='0' then f[i,j]:=0 else f[i,j]:=1;
    end;
    readln;
    end;
    tot:=0;
    for i:=1 to n do
    for j:=1 to m do
    if ((i=1)or(j=1)or(i=n)or(j=m))and(f[i,j]=0) then bfs(i,j);
    for i:=1 to n do
    for j:=1 to m do
    if f[i,j]=0 then inc(tot);
    writeln(tot);
    end.
    第一次零分 第二次满分
    零分 和 满分的差距让我受不了

  • 0
    @ 2015-07-25 23:15:06

    有问题吗???
    const
    way:array[1..4,1..2] of integer=((1,0),(0,1),(-1,0),(0,-1));
    var
    a:array[-1..600,-1..600] of char;
    s,d,f,g,k,sum:longint;
    procedure try(x,y:longint);
    var
    h,j,kk:longint;
    begin
    h:=x; j:=y;
    for kk:=1 to 4 do
    if a[h+way[kk,1],j+way[kk,2]]='0' then begin
    a[h+way[kk,1],j+way[kk,2]]:='*';
    try(h+way[kk,1],j+way[kk,2]);
    end;
    end;
    begin
    readln(s,d);
    for f:=1 to s+1 do begin
    a[0,f]:='0';
    a[d+1,f]:='0';
    end;
    for f:=1 to d+1 do begin
    a[f,0]:='0';
    a[f,s+1]:='0';
    end;
    for f:=1 to s do begin
    for g:=1 to d do read(a[f,g]);
    readln;
    end;
    a[0,0]:='*';
    try(0,0);
    for f:=1 to s do
    for g:=1 to d do
    if a[f,g]='0' then inc(sum);
    writeln(sum);
    end.

  • 0
    @ 2015-07-22 11:49:33

    pascal用到字符串的一定要用**ansi**string!

  • 0
    @ 2015-05-22 23:20:33

    floodfill算法

  • 0
    @ 2014-12-27 17:25:36

    program oibh(input,output);
    var a:array[0..501,0..501] of char;
    sss:char;
    ans,i,j,x,y,k,l,m:longint;
    begin
    readln(y,x);
    for i:=1 to x do
    begin
    for j:=1 to y do
    begin
    read(a[i,j]);
    end;
    readln;
    end;
    for i:=1 to x do
    for j:=1 to y do
    begin
    if(a[i,j]='0')and(a[i-1,j]='*')and(a[i+1,j]='*')and(a[i,j+1]='*')and(a[i,j-1]='*') then inc(ans);
    end;
    write(ans);

    end.
    我的哪里错了,怎么会WA0分?

  • 0
    @ 2014-02-26 16:51:24
  • 0
    @ 2013-10-30 10:46:19

    本人弱菜:请问各位大神我的程序哪里错了,只得了30分,BFS:
    const dx:array[1..4] of integer=(1,0,-1,0);
    dy:array[1..4] of integer=(0,1,0,-1);
    var
    i,j,k,m,n:longint;
    s:string;
    f:array[1..500,1..500] of boolean;
    q:array[1..2,1..250000] of integer;
    procedure sou(x,y:longint);
    var i,xx,yy,t,l:longint;
    begin
    t:=0;l:=1;q[1,1]:=x;q[2,1]:=y;f[x,y]:=false;inc(k);
    while t<=l do
    begin
    inc(t);
    xx:=q[1,t];yy:=q[2,t];
    for i:=1 to 4 do
    if (xx+dx[i]>0)and(xx+dx[i]<=n)and(yy+dy[i]>0)and(yy+dy[i]<=m)
    and(f[xx+dx[i],yy+dy[i]]) then
    begin
    f[xx+dx[i],yy+dy[i]]:=false;
    inc(k);
    inc(l);
    q[1,l]:=xx+dx[i];
    q[2,l]:=yy+dy[i];
    end;
    end;
    end;
    begin
    fillchar(f,sizeof(f),true);
    readln(n,m);
    for i:=1 to n do
    begin
    readln(s);
    for j:=1 to m do
    if s[j]='*' then begin f[i,j]:=false;inc(k); end;
    end;
    for i:=1 to n do
    begin
    if f[i,1] then sou(i,1);
    if f[i,m] then sou(i,m);
    end;
    for i:=1 to m do
    begin
    if f[1,i] then sou(1,i);
    if f[n,i] then sou(n,i);
    end;
    writeln(n*m-k);
    end.

    • @ 2013-10-30 10:50:42

      把s:string改为s:ansistring。。。

  • 0
    @ 2012-11-06 23:11:16

    样例似乎错了?

    暴搜完成。

    编译通过...

    ├ 测试数据 01:答案正确... (0ms, 1680KB)

    ├ 测试数据 02:答案正确... (109ms, 1660KB)

    ├ 测试数据 03:答案正确... (0ms, 1664KB)

    ├ 测试数据 04:答案正确... (0ms, 1688KB)

    ├ 测试数据 05:答案正确... (0ms, 1680KB)

    ├ 测试数据 06:答案正确... (0ms, 1680KB)

    ├ 测试数据 07:答案正确... (0ms, 1672KB)

    ├ 测试数据 08:答案正确... (0ms, 1668KB)

    ├ 测试数据 09:答案正确... (0ms, 1684KB)

    ├ 测试数据 10:答案正确... (0ms, 1672KB)

    …………把m和n反过来读入就过了………………rp不好

  • 0
    @ 2010-04-08 13:28:13

    floodfill,在地图外再加一圈0,从(0,0)开始搜

  • 0
    @ 2009-11-10 11:12:01

    呼 我也非搜索

    program vijos_p1294;

    const

    d:array[1..4,1..2] of longint=((-1,0),(1,0),(0,-1),(0,1));

    max=9;

    var

    map:array[0..501,0..501] of boolean;

    find:array[0..501,0..501] of boolean;

    y,x,iy,ix,num:longint;

    c:char;

    function pd(iy,ix:longint):boolean;

    var i,ty,tx:longint;

    begin

    for i:=1 to 4 do

    begin

    ty:=d+iy;

    tx:=d+ix;

    if ((ty>=1) and (ty=1) and (tx

  • 0
    @ 2009-11-03 22:51:54

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:0ms

    H2O~

    DFS秒杀~

    晒晒程序 半梦半醒间 写得很不好看……

    Const

    dx:array [1..4] of Longint=(1,0,-1, 0);

    dy:array [1..4] of Longint=(0,1, 0,-1);

    Var

    a:array [0..501,0..501] of Char;

    f:array [0..501,0..501] of Boolean;

    i,j,tot,x,y:Longint;

    Procedure try(x1,y1:Longint);

    Var

    i:Longint;

    Begin

    f[x1,y1]:=true;

    For i:=1 to 4 Do

    If (x1+dx[i]=1) Then

    If (y1+dy[i]=1) Then

    If (f[x1+dx[i],y1+dy[i]]=false) and (a[x1+dx[i],y1+dy[i]]='0') Then

    try(x1+dx[i],y1+dy[i]);

    End;

    Begin

    Readln(x,y);

    For i:=1 to x Do Begin

    For j:=1 to y Do Read(a);

    Readln;

    End;

    For i:=1 to x Do If (a='0') and (f=false) Then try(i,1);

    For i:=1 to x Do If (a='0') and (f=false) Then try(i,y);

    For i:=1 to y Do If (a[1,i]='0') and (f[1,i]=false) Then try(1,i);

    For i:=1 to y Do If (a[x,i]='0') and (f[x,i]=false) Then try(x,i);

    tot:=0;

    For i:=1 to x Do

    For j:=1 to y Do If (f=false) and (a='0') Then Inc(tot);

    Writeln(tot);

    End.

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ID
1294
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