题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1175
写了10次,才对。。。
代码都应该很清楚了。一个dfs函数就可以了,参数记录了方向,转弯次数。
代码:
#include <cstdio> #include <cstring> int a[1005][1005]; int v[1005][1005]; int n,m; int x1,y1,x2,y2; int dx[] = {-1,0,1,0}; int dy[] = {0,1,0,-1}; int flag; int isOk(int x,int y,int dir) { int tx = x + dx[dir]; int ty = y + dy[dir]; while(tx>=1&&tx<=n&&ty>=1&&ty<=m) { if(a[tx][ty] == a[x2][y2] && tx == x2 && ty == y2) return 1; else if(a[tx][ty]) return 0; else { tx = tx + dx[dir]; ty = ty + dy[dir]; } } return 0; } void dfs(int x,int y,int cnt,int dir) { // printf("cnt = %d\n",cnt); if(cnt > 2) return ; if(flag) return ; else if(x == x2 && y == y2 && cnt <= 2) { flag = 1; return ; } else if(cnt == 2) { if(isOk(x,y,dir)) flag = 1; } else { for(int i = 0; i < 4; ++i) { int tx = x + dx[i]; int ty = y + dy[i]; if(tx == x2 && ty == y2) { flag = 1; return ; } //printf("cnt = %d i = %d \n",cnt,i); if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&!v[tx][ty]&&a[tx][ty] == 0) { // printf("%d %d %d\n",tx,ty,i); v[tx][ty] = 1; if(i == dir) dfs(tx,ty,cnt,i); else dfs(tx,ty,cnt+1,i); v[tx][ty] = 0; } } } } int main() { while(~scanf("%d%d",&n,&m),n + m) { memset(v,0,sizeof(v)); for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) scanf("%d",&a[i][j]); int q; scanf("%d",&q); while(q--) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); flag = 0; if(!a[x1][y1] || !a[x2][y2] || a[x1][y1] != a[x2][y2]) { printf("NO\n"); continue; } for(int i = 0;i < 4;++i) dfs(x1,y1,0,i); if(flag) printf("YES\n"); else printf("NO\n"); } } }