Codeforces Gym 100187 E. Two Labyrinths (双图BFS找共同最短路)

E. Two Labyrinths

time limit per test
2 seconds

memory limit per test
256 megabytes

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 sizen × 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 thesen 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».

Sample test(s)

Input

3 5
.....
.#.#.
.....

.....
#.#.#
.....

Output

NO

Input

3 5
.....
.#.##
.....

.....
##.#.
.....

Output

YES

题意：给你两个图，问两个图从(0,0)到(n-1,m-1)的最短路径是否相同

思路：用三次BFS，第一次寻找第一个图的最短路，第二次找第二个图的最短路，第三次查看两个图的最短路径是否相同。

这道题让我知道细心是多么重要，WA了10次，在同一组数据上，和傻逼一样查错查了一个半小时，换了几个姿势，最后发现就因为标记数组的大小写错了，就是开数组的时候少按了一个5，一个5，一个5啊，当时找的想撞墙了。。找出来之后简直是想吐血啊啊啊啊啊。。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#define MAXN 1010000
#define LL long long
#define ll __int64
#include<iostream>
#include<algorithm>
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
ll powmod(ll a,LL b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
char map[3][555][555];
int v[3][555][555];//就是这里，写成了v[3][555][55]
int vis[555][555];
int cnt[3];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int n,m;
int bz;
struct s
{
int x,y,step;
};
int check(int kk,int xx,int yy)
{
if(xx<0||xx>=n||yy<0||yy>=m)
return 0;
if(v[kk][xx][yy]||map[kk][xx][yy]=='#')
return 0;
return 1;
}
queue<s>q;
void bfs(int k)
{
while(!q.empty())
q.pop();
s a,b;
a.x=0;a.y=0;a.step=0;
v[k][0][0]=1;
q.push(a);
while(!q.empty())
{
a=q.front();
q.pop();
if(a.x==n-1&&a.y==m-1&&a.step<cnt[k])
{
cnt[k]=a.step;
continue;
}
for(int i=0;i<4;i++)
{
b.x=a.x+dir[i][0];
b.y=a.y+dir[i][1];
if(check(k,b.x,b.y))
{
b.step=a.step+1;
v[k][b.x][b.y]=1;
//printf("debug\n");
q.push(b);
}
}
}
}
void bbfs()
{
while(!q.empty())
q.pop();
s a,b;
a.x=0;a.y=0;a.step=0;
vis[0][0]=1;
q.push(a);
while(!q.empty())
{
if(bz)
break;
a=q.front();
q.pop();
if(a.x==n-1&&a.y==m-1&&a.step==cnt[1])
{
bz=1;
break;
}
for(int 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&&!vis[b.x][b.y]&&map[0][b.x][b.y]=='.'&&map[1][b.x][b.y]=='.')
{
b.step=a.step+1;
vis[b.x][b.y]=1;
q.push(b);
}
}
}
}
int main()
{
int i;
while(scanf("%d%d",&n,&m)!=EOF)
{
mem(map);
for(i=0;i<n;i++)
scanf("%s",map[0][i]);
for(i=0;i<n;i++)
scanf("%s",map[1][i]);
mem(v);
cnt[0]=INF;cnt[1]=INF;
bfs(0);
bfs(1);
//printf("%d %d\n",cnt[0],cnt[1]);
if(cnt[0]!=cnt[1])
{
printf("NO\n");
continue;
}
bz=0;
mem(vis);
bbfs();
if(bz)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}