HDOJ 1175 连连看 (bfs)

来源：互联网发布：计量经济学论文数据编辑：程序博客网时间：2024/06/06 08:39

题目链接：～(￣▽￣～)(～￣▽￣)～

思路：用bfs一次把这个方向上能到的点入队

code：

#include <stdio.h>int m = 0, n = 0, front = 0, rear = 0, map[1002][1002], used[1002][1002];int X1 = 0, Y1 = 0, X2 = 0, Y2 = 0;int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};typedef struct{    int x, y, count;}node;node step, quene[1000*1000+1];int bfs(){    int i = 0, x = 0, y = 0, fx = 0, fy = 0, count = 0;    quene[rear++] = step;    while(front<rear)    {        x = quene[front].x; y = quene[front].y; count = quene[front++].count;        for(i = 0; i<4; i++)        {            fx = x+dir[i][0]; fy = y+dir[i][1];            while(!map[fx][fy] && count<3 && fx>0 && fx<=n && fy>0 &&fy<=m && !used[fx][fy])//这方向上能到的点入队            {                used[fx][fy] = 1;                step.x = fx; step.y = fy; step.count = count+1;                quene[rear++] = step;                fx += dir[i][0]; fy += dir[i][1];            }            if(fx == X2 && fy == Y2 && count<3)                return 1;        }    }    return 0;}int main(){    int i = 0,  j= 0, q = 0;    while(scanf("%d %d",&n, &m) , n && m)    {        for(i = 1; i<=n; i++)            for(j = 1; j<=m; j++)                scanf("%d",&map[i][j]);        scanf("%d",&q);        while(q--)        {            for(i = 1; i<=n; i++)                for(j = 1; j<=m; j++)                    used[i][j] = 0;            scanf("%d %d %d %d",&X1, &Y1, &X2, &Y2);            if((map[X1][Y1] != map[X2][Y2] || !map[X1][Y1] || !map[X2][Y2]) || (X1==X2 && Y2 == Y1))                printf("NO\n");            else            {                front = rear = 0;                step.x = X1; step.y = Y1;step.count = 0;                if(bfs())                    printf("YES\n");                else                    printf("NO\n");            }        }    }    return 0;}