连连看

#include <iostream> #include <queue> #include <cstring> using namespace std; struct node { int x, y; int t, d; }; queue<node> q; int n, m, map[1002][1002], prove; int visit[1002][1002][4]; //记录第（i,j）个点4个方向是否已走过,0为未走过。 int qry, sx, sy, ex, ey; int dx[4] = {0, -1, 0, 1}; int dy[4] = {1, 0, -1, 0}; int check(int x, int y) { if(x < 1 || x > n || y < 1 || y > m) return 0; else return 1; } void BFS() { while(!q.empty()) q.pop(); memset(visit, 0, sizeof(visit)); node s, e; s.x = sx; s.y = sy; s.t = 0; s.d = -1; q.push(s); while(!q.empty()) { s = q.front(); q.pop(); if(s.t > 2) continue; if(s.x == ex && s.y == ey) { prove = 1; cout << "YES" << endl; break; } for(int i = 0; i < 4; i++) { e.x = s.x + dx[i]; e.y = s.y + dy[i]; if(!check(e.x, e.y) || visit[s.x][s.y][i]) //该点该方向是否走过 continue; if( map[e.x][e.y] == 0 || (e.x == ex && e.y == ey) ) { if(s.d == -1 || i == s.d) { e.d = i; e.t = s.t; q.push(e); visit[s.x][s.y][i] = 1; //该点该方向已走过 } else { e.d = i; e.t = s.t + 1; q.push(e); visit[s.x][s.y][i] = 1; } } } } } int main() { while(cin >> n >> m&&n&&m) { for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) cin >> map[i][j]; cin >> qry; for(int i = 1; i <= qry; i++) { cin >> sx >> sy >> ex >> ey; prove = 0; if(map[sx][sy] == map[ex][ey] && map[sx][sy] != 0) BFS(); if(!prove) cout << "NO" << endl; } } return 0; }