HDU 1728 逃离迷宫

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http://acm.hdu.edu.cn/showproblem.php?pid=1728

这道题和 hdu 1175 很相似，所以直接把1175这题的代码那过来修改，可是一直WA，按理说不应该WA的。。。。

后来才发现，在转弯的时候，如果当前的转弯次数和上次相同，这时候也要放到队列里，因为我们不知道那个点能通向终点，这个地方不一样，所以要改为(next.zj <= vis[xx][yy])；哎

ＡＣ代码

// hdu 1728#include <stdio.h>#include <stdlib.h>#include <string.h>#include <queue>using namespace std;int m, n, k, x1, y1, x2, y2, flag;char map[105][105];int vis[105][105];int move[4][2] = {0, -1, 0, 1, -1, 0, 1, 0};struct point{    int x, y, fx, zj;} st;queue<point>q;void bfs();int main(){//freopen("1.in","r",stdin);    int i, j, t;    scanf("%d", &t);    while (t--)    {        scanf("%d %d", &m, &n);        for (i = 1; i <= m; i++)        {            for (j = 1; j <= n; j++)            {                scanf(" %c", &map[i][j]);            }        }        scanf("%d %d %d %d %d",&k, &y1, &x1, &y2, &x2);        for (int ii = 1; ii <= m; ii++)            for (j = 1; j <= n; j++)            {                vis[ii][j] = 1000;            }        st.x = x1;        st.y = y1;        st.fx = -1;        st.zj = 0;        flag = 0;        bfs();        if (flag)        {            printf("yes\n");        }        else        {            printf("no\n");        }    }}void bfs(){    while (!q.empty())    {        q.pop();    }    q.push(st);    while (!q.empty())    {        point now = q.front();        q.pop();//printf("\nnow x %d,y %d, fx %d,zj %d\n", now.x, now.y, now.fx, now.zj);        if (now.x == x2 && now.y == y2 ){flag = 1;return;}        for (int i = 0; i < 4; i++)        {            int xx = now.x, yy = now.y;            xx += move[i][0];            yy += move[i][1];            point next = now;            next.x = xx;            next.y = yy;            next.fx = i;            if (next.fx != now.fx && now.fx != -1)            {                next.zj++;            }            if(next.zj>k) continue;            if (xx > 0 && xx <= m && yy > 0 && yy <= n && map[xx][yy] == '.')            {                if (next.zj <= vis[xx][yy])                {                    vis[xx][yy] = next.zj;                    q.push(next);                }            }        }    }}

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