Gym100187E,Two Labyrinths，广搜

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题意：给你两幅图，判断一下两幅图是否有公共的最短路径，从左上角走到右下角，有的话输出YES，否则输出NO

思路：

首先把两幅图合在一起搜一遍，看是否有公共的路径走到右下角（右下角不一定可达），有的话，记下步数，没有的话，直接输出NO；

然后，搜一遍第一幅图，记下到达右下角的最短步数（如果不可达，第一步之后就结束了）；

再搜一遍第二幅图，记下到达右下角的最短步数；

最后比较这三个步数是否相等，都相等，则输出YES，否则输出NO；

#include<cstdio>#include<iostream>#include<cstring>#include<queue>#include<cstdlib>using namespace std;struct node{    char ch;    int x,y;    int step;};struct node maz1[509][509],maz2[509][509];//第一幅图，第二幅图bool vis[509][509];int n,m;int dir[4][2] = {0,-1,-1,0,0,1,1,0};void BFS()//两幅图合在一起搜一遍{    int i;    struct node a,b;    queue<node>q;    while (!q.empty())        q.pop();    memset(vis,false,sizeof(vis));    maz1[0][0].step = 0;    maz2[0][0].step = 0;    vis[0][0] = true;    q.push(maz1[0][0]);    while (!q.empty())    {        a = q.front();        q.pop();        for (i = 0; i < 4; ++i)        {            b.x = a.x + dir[i][0];            b.y = a.y + dir[i][1];            if (b.x < 0 || b.x >= n || b.y < 0 || b.y >= m)                continue;            if (maz1[b.x][b.y].ch == '.' && maz2[b.x][b.y].ch == '.' && !vis[b.x][b.y])//合在一起            {                b.step = a.step + 1;                maz1[b.x][b.y].step = a.step + 1;                maz2[b.x][b.y].step = a.step + 1;                vis[b.x][b.y] = true;                if (b.x == n-1 && b.y == m-1)                    return ;                q.push(b);            }        }    }    return ;}void BFS1()//搜一遍第一幅图{    int i;    struct node a,b;    queue<node>q;    while (!q.empty())        q.pop();    memset(vis,false,sizeof(vis));    maz1[0][0].step = 0;    q.push(maz1[0][0]);    vis[0][0] = true;    while (!q.empty())    {        a = q.front();        q.pop();        for (i = 0; i < 4; ++i)        {            b.x = a.x + dir[i][0];            b.y = a.y + dir[i][1];            if (b.x < 0 || b.x >= n || b.y < 0 || b.y >= m)                continue;            if (maz1[b.x][b.y].ch == '.' && !vis[b.x][b.y])            {                vis[b.x][b.y] = true;                b.step = a.step + 1;                maz1[b.x][b.y].step = a.step + 1;                if (b.x == n-1 && b.y == m-1)                    return ;                q.push(b);            }        }    }}void BFS2()//搜一遍第二幅图{    int i;    struct node a,b;    queue<node>q;    while (!q.empty())        q.pop();    memset(vis,false,sizeof(vis));    maz2[0][0].step = 0;    q.push(maz2[0][0]);    vis[0][0] = true;    while (!q.empty())    {        a = q.front();        q.pop();        for (i = 0; i < 4; ++i)        {            b.x = a.x + dir[i][0];            b.y = a.y + dir[i][1];            if (b.x < 0 || b.x >= n || b.y < 0 || b.y >= m)                continue;            if (maz2[b.x][b.y].ch == '.' && !vis[b.x][b.y])            {                vis[b.x][b.y] = true;                b.step = a.step + 1;                maz2[b.x][b.y].step = a.step + 1;                if (b.x == n-1 && b.y == m-1)                    return ;                q.push(b);            }        }    }}int main(){    int i,j;    while (~scanf("%d%d",&n,&m))    {        for (i = 0; i < n; ++i)        {            for (j = 0; j < m; ++j)            {                cin>>maz1[i][j].ch;                maz1[i][j].x = i;                maz1[i][j].y = j;            }        }//存第一幅图        for (i = 0; i < n; ++i)        {            for (j = 0; j < m; ++j)            {                cin>>maz2[i][j].ch;                maz2[i][j].x = i;                maz2[i][j].y = j;            }        }//存第二幅图        maz1[n-1][m-1].step = -1;        maz2[n-1][m-1].step = -1;//刚开始把两幅图的终点步数都标记成-1        BFS();        if (maz1[n-1][m-1].step == -1 || maz2[n-1][m-1].step == -1)//搜完之后，如果终点步数还是-1的话，说明没有公共路径可达终点            printf("NO\n");        else        {            int ans = maz1[n-1][m-1].step;//记录下刚才搜到的结果            BFS1();            BFS2();            if (ans == maz1[n-1][m-1].step && ans == maz2[n-1][m-1].step)//比较这三个步数是否都相等            {                printf("YES\n");            }            else            {                printf("NO\n");            }        }    }    return 0;}

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