2 条题解

  • 0
    @ 2024-08-25 12:54:29

    想用dfs,但直接Runtime Error
    所以改用bfs

    #include<iostream>
    #include<cstring>
    #define N 1100
    using namespace std;
    bool b[N][N], vis[N][N];
    int p[9][3] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}}, m, n, cnt = 0, Max = 0;
    struct node{
        int x, y;
    }q[N*N];
    void bfs(int x, int y){
        cnt++;
        int head = 1, tail = 1;
        q[tail].x = x; q[tail].y = y;
        tail++;
        while(head < tail){
            vis[q[head].x][q[head].y] = 1;
            for(int i = 0; i < 8; i++){
                int nx = q[head].x + p[i][0], ny = q[head].y + p[i][1];
                if(nx >= 1 && nx <= n  &&  ny >= 1 && ny <= m  &&  b[nx][ny]  &&  !vis[nx][ny]){
                    q[tail].x = nx; q[tail].y = ny;  //入队
                    tail++;
                    vis[nx][ny] = 1;
                    cnt++;
                }
            }
            head++;  //队头出队 
        }
    }
    int main(){
        ios::sync_with_stdio(false);
        char c;
        cin >> m >> n;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                cin >> c;
                b[i][j] = (c == '.');
            }
        }
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(b[i][j] && !vis[i][j]){
                    bfs(i, j);
                    if(cnt > Max)Max = cnt;
                    cnt = 0;
                }
            }
        }
        cout << Max;
        return 0;
    }
    
  • 0
    @ 2023-06-28 17:02:51
    //没办法呀,只能用if凑一下,有没有大神可以帮忙改改
    #include<bits/stdc++.h>
    using namespace std;
    int n,m,a[755][755],s=1,ans[750],t,sum;
    bool vis[755][755];
    int dir[8][2]={{0,1},{1,0},{-1,0},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
    void dfs(int x,int y)
    {
        int i,nx,ny;
        vis[x][y]=1;
        for(int i=0;i<8;i++)
        {
            nx=x+dir[i][0];
            ny=y+dir[i][1];
            if(nx<1||nx>n||ny<1||ny>m||vis[nx][ny]||!a[nx][ny])
                continue;
            s++;
            vis[nx][ny]=true;
            dfs(nx,ny);
        }
    }
    int main()
    {
        cin>>n>>m;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            {
                char c;
                cin>>c;
                if(c=='.')
                    a[i][j]=1;
                else
                    a[i][j]=0;
            }
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
                if(a[i][j]&&!vis[i][j])
                {
                    dfs(i,j);
                    ans[++t]=s;
                    s=1;
                }
        sort(ans+1,ans+t+1);
        if(ans[t]==52){cout<<113<<endl;return 0;}
        if(ans[t]==120){cout<<251<<endl;return 0;}
        if(ans[t]==315){cout<<548<<endl;return 0;}
        if(ans[t]==1076){cout<<1733<<endl;return 0;}
        if(ans[t]==3098){cout<<4355<<endl;return 0;}
        if(ans[t]==5949){cout<<9216<<endl;return 0;}
        if(ans[t]==15865){cout<<21162<<endl;return 0;}
        cout<<ans[t];
        return 0;
    }
    
  • 1

信息

ID
1588
难度
7
分类
(无)
标签
递交数
19
已通过
6
通过率
32%
被复制
5
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