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  • 百度之星初赛A hdu6113

    度度熊的01世界

    题意:中文题

    思路:在最外面加一圈0,然后判联通块的个数,1的联通块数量只能是1,否则输出-1,0的联通块只能是1或者2,否则输出-1

    AC代码:

    #include "iostream"
    #include "string.h"
    #include "stack"
    #include "queue"
    #include "string"
    #include "vector"
    #include "set"
    #include "map"
    #include "algorithm"
    #include "stdio.h"
    #include "math.h"
    #pragma comment(linker, "/STACK:102400000,102400000")
    #define ll long long
    #define endl ("
    ")
    #define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
    #define mem(a,x) memset(a,x,sizeof(a))
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define ft first
    #define sd second
    #define lrt (rt<<1)
    #define rrt (rt<<1|1)
    using namespace std;
    const long long INF = 1e18+1LL;
    const int inf = 1e9+1e8;
    const int N=1e5+100;
    const ll mod=1e9+7;
    
    int n,m;
    int G[105][105],vis[2][105][105];
    int d[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
    struct Node{
        int xx, yy;
    };
    void bfs(int x, int y, int k){
        queue<Node> Q;
        while(!Q.empty()) Q.pop();
        Node now,next; now.xx=x, now.yy=y;
        Q.push(now), vis[k][x][y]=1;
        while(!Q.empty()){
            now=Q.front(); Q.pop();
            for(int i=0; i<4; ++i){
                next.xx=now.xx+d[i][0];
                next.yy=now.yy+d[i][1];
                if(next.xx>n+1 || next.xx<0 || next.yy>m+1 || next.yy<0) continue;
                if(G[next.xx][next.yy]==k && !vis[k][next.xx][next.yy]){
                    Q.push(next);
                    vis[k][next.xx][next.yy]=1;
                }
            }
        }
    }
    int main(){
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        while(cin>>n>>m){
            mem(G,0), mem(vis,0);
            for(int i=1; i<=n; ++i){
                for(int j=1; j<=m; ++j){
                    char cc; cin>>cc;
                    G[i][j]=cc-'0';
                }
            }
            int c0=0, c1=0;
            for(int i=0; i<=n+1; ++i){
                for(int j=0; j<=m+1; ++j){
                    if(G[i][j]==0 && !vis[0][i][j]){
                        bfs(i,j,0); c0++;
                    }
                    else if(G[i][j]==1 && !vis[1][i][j]){
                        bfs(i,j,1); c1++;
                    }
                }
            }
            if(c1!=1 || c0>=3) cout<<-1<<endl;
            else if(c0==1) cout<<1<<endl;
            else cout<<0<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/max88888888/p/7351596.html
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