题解

175 条题解

  • 3
    @ 2017-11-04 19:54:54

    数据太水,四重循环暴力解答 (o ° ω ° O )

    #include <iostream>
    using namespace std;
    char a[10005][10005];
    int n,m;
    int main()
    {
        int ans=0,tot;
        cin>>m>>n;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>a[i][j];
            }
        }            
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=m;j++)
            {
                tot=0;
                for(int q=i;q<=i+n-1;q++)
                {
                    for(int p=j;p<=j+n-1;p++)
                    {                
                        if(a[q][p]=='#')
                        tot++;
                        if(tot>ans)
                        ans=tot;
                    }
                }
            }
        }
        cout<<ans<<endl;
        return 0;
    }
    
  • 2
    @ 2018-08-08 11:53:33
    /*pwq*/
    #include<cstdio> 
    #include<cstring>
    using namespace std;
    int a[1100][1100],k;
    char tmp[1100];
    int chk(int x,int y)
    {
        int ret=0;
        for(int i=x;i<x+k;i++)
        {
            for(int j=y;j<y+k;j++)
            {
                if(a[i][j])
                   ret++;
            }
        }
        return ret;
    }
    int main()
    {
        int n,max=0;
        scanf("%d %d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",tmp);
            for(int j=1;j<=n;j++)
            {
                if(tmp[j-1]=='#')
                    a[i][j]=1;
            }
        }
        for(int i=1;i<=n-k+1;i++)
        {
            for(int j=1;j<=n-k+1;j++)
            {
                int tmp=chk(i,j);
                if(max<tmp)
                    max=tmp;
            }
        }
        printf("%d",max);
        return 0;  
    }  
    
    
  • 1
    @ 2019-08-17 21:13:50

    数据很水?其实可以做前缀和,速度快一些。

    #include <iostream>
    
    using namespace std;
    
    int m,n;
    bool jd[10000][10000]={0};
    
    int main()
    {
        cin>>m>>n;
        int i,j,k,now,ans=0;
        char c;
        for(i=0;i<m;i++)
        {
            for(j=0;j<m;j++)
            {
                cin>>c;
                if(c=='#')
                {
                    jd[i][j]=true;
                }
            }
        }
        for(i=m-n;i>=0;i--)
        {
            now=0;
            for(j=0;j<n;j++)
            {
                for(k=0;k<n;k++)
                {
                    if(jd[i+j][m-n+k])
                    {
                        now++;
                    }
                }
            }
            ans=max(ans,now);
            for(j=m-n-1;j>=0;j--)
            {
                for(k=0;k<n;k++)
                {
                    if(jd[i+k][j])
                    {
                        now++;
                    }
                    if(jd[i+k][j+n])
                    {
                        now--;
                    }
                }
                ans=max(ans,now);
            }
        }
        cout<<ans<<endl;
        return 0;
    }
    
  • 1
    @ 2018-09-23 17:32:24

    四重循环不超时……

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <algorithm>
    using namespace std;
    
    int main()
    {
        int m,n;
        scanf("%d %d",&m,&n);
        getchar();
        char d[m][m];
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<m;j++)
                d[i][j]=getchar();
            getchar();
        }
        int ans=-1;
        for(int i=0;i<m-n+1;i++)
            for(int j=0;j<m-n+1;j++)
            {
                int sum=0;
                for(int p=i;p<n+i;p++)
                    for(int q=j;q<n+j;q++)
                        if(d[p][q]=='#')
                            sum++;
                ans=max(ans,sum);
            }
        cout<<ans<<endl;
        return 0;
    }
    
  • 1
    @ 2018-05-06 17:37:25

    So Water
    ```cpp
    #include<cstdlib>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int n,m;
    int a[300][300];
    int main()
    {
    int max=0;
    memset(a,0,sizeof(a));
    cin>>m>>n;
    for(int i=1;i<=m;i++)
    {
    getchar();
    for(int j=1;j<=m;j++)
    {
    char c;
    c=getchar();
    if(c=='#')a[i][j]=a[i-1][j]+a[i][j-1]-a[i-1][j-1]+1;
    else a[i][j]=a[i-1][j]+a[i][j-1]-a[i-1][j-1];
    }
    }
    for(int i=1;i<=m;i++)
    {
    for(int j=1;j<=m;j++)

    {

    int nmax,vi,vj;
    vi=i+n-1;if(vi>m)vi=m;
    vj=j+n-1;if(vj>m)vj=m;
    nmax=a[vi][vj]-a[i-1][vj]-a[vi][j-1]+a[i-1][j-1];
    if(nmax>max)max=nmax;
    }
    }
    if(n==0)cout<<0;
    else cout<<max;
    return 0;
    }

  • 1
    @ 2018-05-06 17:35:15

    容斥原理
    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<string>
    #include<stdlib.h>
    using namespace std;
    int m,n;
    int f[1200][1200],sum,sumB[120][120],ans=0;

    int main(){
    cin >> m;
    cin >> n;
    if (n==0){cout << 0;return 0;}
    memset(f,0,sizeof(f));
    memset(sumB,0,sizeof(sumB));
    for (int i=1;i<=m;i++)
    for (int j=1;j<=m;j++){
    char a;
    cin >> a;
    if (a=='#')f[i][j]=1;
    else if (a=='.')f[i][j]=0;
    if (j==m)getchar();
    }
    for (int right=1;right<=m;right++)sumB[0][right]=0;
    for (int bottom=1;bottom<=m;bottom++){
    sum=0;
    for (int right=1;right<=m;right++){
    sum+=f[bottom][right];
    sumB[bottom][right]=sumB[bottom-1][right]+sum;
    }
    }
    for (int i=1;i<=m-n+1;i++)
    for (int j=1;j<=m-n+1;j++){
    int b=sumB[i+n-1][j+n-1]-sumB[i+n-1][j-1]-sumB[i-1][j+n-1]+sumB[i-1][j-1];
    if (b>ans)ans=b;
    }
    cout << ans;
    return 0;
    }

  • 1
    @ 2017-11-02 20:19:34

    前缀和+枚举核弹头位置

    #include<ctime>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n,m;
    int a[11234][11234];
    int main()
    {
        int i,j;
        cin>>n>>m;
        if(m==0){
            cout<<"0";
                return 0;
        }
        else{
            
            for(i=1;i<=n;i++){
                for(j=1;j<=n;j++){
                    char ch;
                    cin>>ch;
                    if(ch=='#'){
                        a[i][j]++;
                    
                    }
                    a[i][j]+=a[i-1][j]+a[i][j-1]-a[i-1][j-1];
                }
            }
        }           
        int ans=0;
        int x1=0,y1=0;
        for(i=1;i<=n-m+1;i++){
            for(j=1;j<=n-m+1;j++){
                //枚举弹头
                int t=a[i+m-1][j+m-1]-a[i-1][j+m-1]-a[i+m-1][j-1]+a[i-1][j-1];
                
                ans=max(ans,t);
        
                }
                
            }
    cout<<ans<<endl;
        return 0;
    }
    
  • 1
    @ 2017-10-05 00:53:48

    so water

    #include <cstdio>
    #include <cstdlib>
    #include <algorithm>
    #include <iostream>  
    #include <cmath>  
    #include <cstring>  
    #include <set>  
    using namespace std;
    int a[11000][11000],k;
    int check(int x,int y)
    {
        int all=0;
        for(int i=x;i<x+k;i++)
        {
            for(int j=y;j<y+k;j++)
            {
                if(a[i][j])
                    all++;
            }
        }
        return all;
    }
    int main()
    {
        int n,max=0;
        scanf("%d %d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            char line[11000];
            scanf("%s",line);
            for(int j=1;j<=n;j++)
            {
                if(line[j-1]=='#')
                    a[i][j]=1;
            }
        }
        for(int i=1;i<=n-k+1;i++)
        {
            for(int j=1;j<=n-k+1;j++)
            {
                int p=check(i,j);
                if(max<p)
                    max=p;
            }
        }
        printf("%d",max);
        return 0;  
    }  
    
  • 1
    @ 2016-12-07 10:36:27
    #include <bits/stdc++.h>
    using namespace std;
    int ans[10001][10001];
    char a; 
    int main()
    {
        int max=0,i,j,k,l,n,m;
        memset(ans,0,sizeof(ans));
        cin>>n;
        cin>>m;
        for(i=1;i<=n;i++) {
            for(j=1;j<=n;j++) {
                cin>>a;
                if(a=='#') ans[i][j]=1;
            }
        }
        for(i=1;i<=n;i++) {
            for(j=1;j<=n;j++) {
                k=0;
                for(int h=i;h<i+m;h++) {
                    for(int hh=j;hh<j+m;hh++) {
                        k+=ans[h][hh];
                    }
                }
                if(k==m*m) {
                    cout<<k<<endl;
                    return 0;
                }
                if(k>max) max=k;
            }
        }
        cout<<max<<endl;
        return 0;
    }
    
  • 0
    @ 2016-10-17 19:31:34

    需要剪枝?

  • 0
    @ 2016-10-06 19:48:45

    Free Pascal Compiler version 3.0.0 [2015/11/16] for i386
    Copyright (c) 1993-2015 by Florian Klaempfl and others
    Target OS: Win32 for i386
    Compiling foo.pas
    Linking foo.exe
    24 lines compiled, 0.0 sec, 28016 bytes code, 1268 bytes data

    测试数据 #0: Accepted, time = 0 ms, mem = 98656 KiB, score = 20

    测试数据 #1: Accepted, time = 0 ms, mem = 98656 KiB, score = 20

    测试数据 #2: Accepted, time = 0 ms, mem = 98656 KiB, score = 20

    测试数据 #3: Accepted, time = 0 ms, mem = 98656 KiB, score = 20

    测试数据 #4: Accepted, time = 0 ms, mem = 98656 KiB, score = 20

    Accepted, time = 0 ms, mem = 98656 KiB, score = 100

  • 0
    @ 2016-09-02 20:55:19

    暴力搜索
    世界核平
    #include <cstdio>
    #include <cstring>

    int main(){
    freopen("in.txt","r",stdin);
    int m,n;
    scanf("%d%d",&m,&n);
    char A[1000][1000];
    memset(A,'.',sizeof(A));
    for(int i=0;i<m;i++)
    for(int j=0;j<m;j++)
    scanf(" %c",&A[i][j]);
    int cnt,ans=0;
    for(int i=0;i<m;i++)
    for(int j=0;j<m;j++){
    cnt=0;
    for(int ai=0;ai<n;ai++)
    for(int aj=0;aj<n;aj++)
    if(A[i+ai][j+aj]=='#')
    cnt++;
    ans=ans>cnt?ans:cnt;
    }
    printf("%d",ans);
    return 0;
    }

  • 0
    @ 2016-08-20 22:13:00

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int n,m,k=0,sum=0;
    char a[1001];
    int bome[1001][1001]={0},qu,su;
    int ms(int ,int );

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

    if(n>m)

    n=m;
    for(int i=1;i<=m;i++)
    for(int j=1;j<=m;j++)
    {if(k>m*m)break;
    k++;
    if(a[k]=='#'){bome[i][j]=bome[i][j-1]+1;}
    else {bome[i][j]=bome[i][j-1];}

    }

    for(int i=1;i<=m-n+1;i++)
    for(int j=1;j<=m-n+1;j++)
    if(sum<ms(i,j))
    sum=ms(i,j);

    cout<<sum;

    return 0;
    }

    int ms(int q,int s )
    {

    int muqian=0;
    for(int i=1;i<=n;i++)
    {

    muqian+=bome[q][s+n-1]-bome[q][s-1];

    q++;
    }
    return muqian;

    }

  • 0
    @ 2016-07-22 13:37:54

    大水题,无力吐槽,暴搜直接过
    pascal
    program dsfa;
    var a:array[1..10000,1..10000] of char;
    i,j,k,l,n,m,s,t,v:longint;
    begin
    readln(m);
    readln(n);
    s:=0;
    for i:=1 to m do
    begin
    for j:=1 to m do
    read(a[i,j]);
    readln;
    end;
    for i:=1 to m-n+1 do
    for j:=1 to m-n+1 do
    begin
    t:=0;
    for k:=i to i+n-1 do
    for l:=j to j+n-1 do
    if a[k,l]='#' then t:=t+1;
    if t>s then s:=t;
    end;
    writeln(s);
    end.

  • 0
    @ 2016-07-18 20:56:35

    一个一个读字符需readln 否则回车被读入 答案就错了

  • 0
    @ 2016-06-14 13:28:21
    记录信息
    评测状态    Accepted
    题目  P1199 核弹危机
    递交时间    2016-06-14 13:28:29
    代码语言    C++
    评测机 ShadowShore
    消耗时间    0 ms
    消耗内存    98372 KiB
    评测时间    2016-06-14 13:28:30
    评测结果
    编译成功
    
    测试数据 #0: Accepted, time = 0 ms, mem = 98372 KiB, score = 20
    测试数据 #1: Accepted, time = 0 ms, mem = 98364 KiB, score = 20
    测试数据 #2: Accepted, time = 0 ms, mem = 98372 KiB, score = 20
    测试数据 #3: Accepted, time = 0 ms, mem = 98368 KiB, score = 20
    测试数据 #4: Accepted, time = 0 ms, mem = 98368 KiB, score = 20
    Accepted, time = 0 ms, mem = 98372 KiB, score = 100
    代码
    #include<cstdio>
    using namespace std;
    const int MAXN=10001;
    struct house
    {
      char a[MAXN];
    }map[MAXN];
    int m,n,ans=0;
    int main()
    {
      //freopen("crisis.in","r",stdin);
      //freopen("crisis.out","w",stdout);
      scanf("%d\n%d",&m,&n);
      for(int i=1;i<=m;i++)
        scanf("%s",&map[i].a[1]);
      for(int i=1;i<=m-n+1;i++)
        for(int j=1;j<=m-n+1;j++)
        {
          int count=0;
          for(int k=i;k<=i+n-1;k++)
            for(int t=j;t<=j+n-1;t++)
              if(map[k].a[t]=='#')count++;
          if(count>ans)ans=count;
        }
      printf("%d",ans);
      return 0;
    }
    
  • 0
    @ 2016-02-01 14:12:46

    数据小,没想到暴力也能过
    #include<iostream>
    using namespace std;

    char s[10001][10001];
    int n,m;
    int main()
    {
    int max=0;
    cin>>n>>m;
    for (int i=1;i<=n;i++)
    for (int j=1;j<=n;j++)
    cin>>s[i][j];
    int t;
    for (int i=1;i<=n-m+1;i++)
    for (int j=1;j<=n-m+1;j++)
    {
    t=0;
    for (int k=i;k<=i+m-1;k++)
    for (int l=j;l<=j+m-1;l++)
    if (s[k][l]=='#') t++;
    if (max<t)max=t;
    }
    cout<<max;
    }

  • 0
    @ 2015-10-06 10:21:58

    #include<cstdio>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<string>
    #include<queue>
    #include<cmath>
    using namespace std;
    int n,m;
    char yu[10000][10000];
    int d[10000][10000];
    int ans(int x,int y)
    {
    int sum=0;
    for(int i=y;i<y+m;i++)
    {
    sum+=d[x][i]-d[x+m][i];
    }
    return sum;
    }
    int main()
    {
    cin>>n>>m;
    for(int i=0;i<n;i++)
    {
    for(int k=0;k<n;k++)
    {
    cin>>yu[i][k];
    }
    }int sum;
    for(int i=0;i<n;i++)
    {
    sum=0;
    for(int k=n-1;k>=0;k--)
    {
    if(yu[i][k]=='#')sum++;
    d[k][i]=sum;
    }
    }
    int tt=0;
    for(int i=0;i<n;i++)
    {
    for(int k=0;k<n;k++)
    {
    tt=max(ans(i,k),tt);
    }
    }
    cout<<tt;
    return 0;
    }

  • 0
    @ 2015-10-06 06:19:16

    #include<cstdio>
    #include<cmath>
    #include<vector>
    using namespace std;

    unsigned short m,n;
    char c;
    unsigned int ans=0;
    vector<vector<unsigned int> > map;

    int main(){
    scanf("%hu\n%hu\n",&n,&m);
    map.resize(n+1);
    for(vector<vector<unsigned int> >::iterator i=map.begin();i!=map.end();i++){
    i->resize(n+1);
    }

    for(int i=1;i<=n;i++){
    for(int j=1;j<=n;j++){
    scanf("%c",&c);
    map[i][j]=map[i-1][j]+map[i][j-1]-map[i-1][j-1]+(c=='#');
    if(i>=m&&j>=m){
    ans=max(ans,map[i][j]-map[i-m][j]-map[i][j-m]+map[i-m][j-m]);
    }
    }
    scanf("\n");
    }

    if(m>n){
    ans=map[n][n];
    }
    printf("%u\n",ans);
    }
    这数据神水啊,这样用的STL都秒过!
    测试数据 #0: Accepted, time = 0 ms, mem = 516 KiB, score = 20
    测试数据 #1: Accepted, time = 0 ms, mem = 516 KiB, score = 20
    测试数据 #2: Accepted, time = 1 ms, mem = 520 KiB, score = 20
    测试数据 #3: Accepted, time = 1 ms, mem = 516 KiB, score = 20
    测试数据 #4: Accepted, time = 0 ms, mem = 520 KiB, score = 20

    • @ 2016-08-21 22:02:23

      其实这样用stl应该不会多费多少时间的

  • 0
    @ 2015-10-06 00:03:26

    测试数据 #0: Accepted, time = 0 ms, mem = 520 KiB, score = 20
    测试数据 #1: WrongAnswer, time = 0 ms, mem = 516 KiB, score = 0
    测试数据 #2: Accepted, time = 0 ms, mem = 516 KiB, score = 20
    测试数据 #3: WrongAnswer, time = 0 ms, mem = 512 KiB, score = 0
    测试数据 #4: Accepted, time = 1 ms, mem = 516 KiB, score = 20
    WrongAnswer, time = 1 ms, mem = 520 KiB, score = 60
    代码
    #include <iostream>
    #include <vector>
    #include <cstdio>

    using namespace std;

    int main()
    {
    long long int n,m;
    vector<char>mapp;
    vector< vector<char> >maap;
    cin>>n>>m;
    long long int i,j;
    int fh=0;
    for(j=0;j<n;++j)
    {
    mapp.push_back('a');
    }
    for(i=0;i<n;++i)
    {
    for(j=0;j<n;++j)
    {
    cin>>mapp[j];
    if(mapp[j]=='#')fh++;
    }
    maap.push_back(mapp);
    }
    if(m>=n)
    {
    cout<<fh;
    return 0;
    }
    static long long int x,y;
    long long int i1,j1;
    static long long int ans=-1;
    static long long int w;
    for(i=0;y<n;++i)
    {
    y=i+m;
    for(j=0;x<n;++j)
    {
    x=j+m;
    for(i1=i;i1<y;++i1)
    {
    for(j1=j;j1<x;++j1)
    {
    if(maap[i1][j1]=='#')w++;
    }
    }
    ans=ans>w? ans:w;
    w=0;
    }
    }
    cout<<ans;
    return 0;
    }
    无奈了。。。。。

信息

ID
1199
难度
4
分类
搜索 | 搜索与剪枝 点击显示
标签
递交数
2704
已通过
1224
通过率
45%
被复制
4
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