HDU5335——贪心+BFS——Walk Out

/*
题意：从左上到右下要求二进制最小，删去前置0
贪心+bfs
先用bfs求出前置0的长度
再以距离为状态进行贪心
*/
#include <bits/stdc++.h>
using namespace std;

int dirx[] = {1, -1, 0, 0};
int diry[] = {0, 0, 1, -1};
const int MAX = 1000 + 10;
char a[MAX][MAX];
int vis[MAX][MAX];
int b[MAX][MAX];
typedef struct {
    int x, y;
}P;
queue <P> q;
int n, m;

int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
    scanf("%d%d", &n, &m);
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= n; i++)
        scanf("%s", a[i]+1);
    for(int i = 0; i <= n; i++)
        b[i][0] = b[i][m+1] = 2;
    for(int i = 0; i <= m; i++)
        b[0][i] = b[n+1][i] = 2;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            b[i][j] = a[i][j]-‘0‘;
        }
    }
    vis[1][1] = 1;
    if(b[1][1] == 0){
        while(!q.empty()) q.pop();
        q.push((P){1, 1});
        while(!q.empty()){
            P now = q.front();
            q.pop();
            int x = now.x;
            int y = now.y;
            for(int i = 0 ; i< 4; i++){
                int X = x + dirx[i], Y = y + diry[i];
                if(X >= 1 && X <= n && Y >= 1 && Y <= m && !vis[X][Y] && b[X][Y] == 0){
                    vis[X][Y] = 1;
                    q.push((P){X, Y});
                }
            }
        }
    }
   // for(int i = 1; i <= n; i++){
   //     for(int j = 1; j <= m; j++)
   //         printf("%d",vis[i][j]);
   //     puts("");
   // }
    if(vis[n][m] && b[n][m] == 0) {
        printf("0\n"); continue;
    }
    if(b[1][1] == 1) printf("1");
    int max_dis = 2;
     for(int i = 1; i <= n ; i++)
        for(int j = 1; j <= m; j++)
            if(vis[i][j])
              max_dis = max(max_dis, i+j);
    for(int i = max_dis; i < n + m; i++){
        int min1 = 1;
        for(int j = 1; j <= n; j++)
            if(i - j >= 1 && i - j <= m && vis[j][i-j])
                min1 = min(min1,min(b[j+1][i-j], b[j][i-j+1]));
                    printf("%d", min1);
        for(int j = 1; j <= n; j++)
            if(i - j >= 1 && i - j <= m && vis[j][i-j]){
                if(b[j+1][i-j] == min1) vis[j+1][i-j] = 1;
                if(b[j][i-j+1] == min1) vis[j][i-j+1] = 1;
            }
    }
    printf("\n");
    }
    return 0;
}