E. Two Labyrinths
time limit per test
2.0 s
memory limit per test
256 MB
input
standard input
output
standard output
A labyrinth is the rectangular grid, each of the cells of which is either free or wall, and it's possible to move only between free cells sharing a side.
Constantine and Mike are the world leaders of composing the labyrinths. Each of them has just composed one labyrinth of size n × m, and now they are blaming each other for the plagiarism. They consider that the plagiarism takes place if there exists such a path from the upper-left cell to the lower-right cell that is the shortest for both labyrinths. Resolve their conflict and say if the plagiarism took place.
Input
In the first line two integers n and m (1 ≤ n, m ≤ 500) are written — the height and the width of the labyrinths.
In the next n lines the labyrinth composed by Constantine is written. Each of these n lines consists of m characters. Each character is equal either to «#», which denotes a wall, or to «.», which denotes a free cell.
The next line is empty, and in the next n lines the labyrinth composed by Mike is written in the same format. It is guaranteed that the upper-left and the lower-right cells of both labyrinths are free.
Output
Output «YES» if there exists such a path from the upper-left to the lower-right cell that is the shortest for both labyrinths. Otherwise output «NO».
Examples
input
Copy
3 5 ..... .#.#. ..... ..... #.#.# .....
output
Copy
NO
input
Copy
3 5 ..... .#.## ..... ..... ##.#. .....
output
Copy
YES
题目连接：
http://codeforces.com/gym/100187/problem/E
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
char map[2][510][510];//存两个地图
int visit[2][510][510];//为防止破坏map地图而设置的记录方位的路径变
int cnt[2];//统计两张地图的最短步数
int dirx[4]={-1,1,0,0};
int diry[4]={0,0,-1,1};
int n,m;
int flag;
struct Node
{
int x,y,step;
};
queue<Node> q;
int cheak(struct Node x,int k)
{
if(x.x>=0&&x.x<n&&x.y>=0&&x.y<m&&(map[k][x.x][x.y]=='.'&&visit[k][x.x][x.y]==0))
{
return 1;
}
else
{
return 0;
}
}
void bfs(int k) //该题是一定存在最短路的
{
while(!q.empty())
{
q.pop();
}
visit[k][0][0]=1;
Node a,b;
a.x=0,a.y=0,a.step=1;
q.push(a);
while(!q.empty())
{
a=q.front();
q.pop();
if(a.x==n-1&&a.y==m-1)
{
cnt[k]=min(cnt[k],a.step);
}
for(int i=0;i<4;i++)
{
b.x=a.x+dirx[i];
b.y=a.y+diry[i];
if(cheak(b,k))//判断下一个点是否可走
{
b.step=a.step+1;
visit[k][b.x][b.y]=1;
q.push(b);
}
}
}
}
void bbfs(int k) //走两个图的bfs，妙
{
while(!q.empty())
{
q.pop();
} //清空栈的操作
visit[k][0][0]=1;
Node a,b;
a.x=0,a.y=0,a.step=1;
q.push(a);
while(!q.empty())
{
a=q.front();
q.pop();
if(a.x==n-1&&a.y==m-1)
{
if(cnt[k]==a.step)
{
flag=1;
break;//广搜的特点是一旦到达目的地，那么一定是最短路
}
}
for(int i=0;i<4;i++)
{
b.x=a.x+dirx[i];
b.y=a.y+diry[i];
if(cheak(b,k)&&map[0][b.x][b.y]=='.'&&map[1][b.x][b.y]=='.')//两个图都能走,妙
{
b.step=a.step+1;
visit[k][b.x][b.y]=1;
q.push(b);
}
}
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%s",map[0][i]);
}
for(int i=0;i<n;i++)
{
scanf("%s",map[1][i]);
}
cnt[0]=0x3f3f3f3f;
cnt[1]=0x3f3f3f3f;
memset(visit,0,sizeof(visit));
bfs(0);
bfs(1);
if(cnt[0]!=cnt[1])//如果各自的最短路的步数都不一样，那么肯定不是一张抄袭的图
{
printf("NO\n");
return 0;
}
memset(visit,0,sizeof(visit));
flag=0;
bbfs(0);//两张图一起搜
if(flag==1)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
return 0;
}