E. Two Labyrinths
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100187/problem/E
Description
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»
Sample Input
3 5
.....
.#.#.
.....
.....
#.#.#
.....
Sample Output
YES
HINT
题意
问你是否存在一条路，在两个迷宫都合法，而且都是最短路呢？
题解：
3次bfs，第一次bfs找到第一个迷宫的最短路，第二次bfs找到第二个迷宫的最短路，第三次bfs求共同的最短路
代码
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** int a1[][];
int a2[][];
string s;
struct node
{
int x,y,z;
};
int dx[]={,-,,};
int dy[]={,,,-};
int vis[][];
int main()
{
int n=read(),m=read();
for(int i=;i<n;i++)
{
cin>>s;
for(int j=;j<m;j++)
{
if(s[j]=='.')
a1[i][j]=;
else
a1[i][j]=;
}
}
for(int i=;i<n;i++)
{
cin>>s;
for(int j=;j<m;j++)
{
if(s[j]=='.')
a2[i][j]=;
else
a2[i][j]=;
}
}
queue<node> q;
q.push((node){,,});
int flag=;
int ans1=inf;
memset(vis,,sizeof(vis));
vis[][]=;
while(!q.empty())
{
node now=q.front();
q.pop();
if(now.x==n-&&now.y==m-&&ans1>now.z)
{
ans1=now.z;
continue;
}
for(int i=;i<;i++)
{
node next;
next.x=now.x+dx[i];
next.y=now.y+dy[i];
next.z=now.z;
if(next.x<||next.x>=n)
continue;
if(next.y<||next.y>=m)
continue;
if(vis[next.x][next.y]||a1[next.x][next.y]==)
continue;
vis[next.x][next.y]=;
q.push((node){next.x,next.y,next.z+});
}
}
while(!q.empty())
q.pop();
q.push((node){,,});
int ans2=inf;
memset(vis,,sizeof(vis));
vis[][]=;
while(!q.empty())
{
node now=q.front();
q.pop();
if(now.x==n-&&now.y==m-&&ans2>now.z)
{
ans2=now.z;
continue;
}
for(int i=;i<;i++)
{
node next;
next.x=now.x+dx[i];
next.y=now.y+dy[i];
next.z=now.z;
if(next.x<||next.x>=n)
continue;
if(next.y<||next.y>=m)
continue;
if(vis[next.x][next.y]||a2[next.x][next.y]==)
continue;
vis[next.x][next.y]=;
q.push((node){next.x,next.y,next.z+});
}
} if(ans1!=ans2)
{
puts("NO");
return ;
}
while(!q.empty())
q.pop();
q.push((node){,,});
memset(vis,,sizeof(vis));
vis[][]=; while(!q.empty())
{
if(flag)
break;
node now=q.front();
q.pop();
if(now.x==n-&&now.y==m-&&now.z==ans1)
flag=;
if(flag)
break;
for(int i=;i<;i++)
{
node next;
next.x=now.x+dx[i];
next.y=now.y+dy[i];
next.z=now.z+;
if(next.x<||next.x>=n)
continue;
if(next.y<||next.y>=m)
continue;
if(vis[next.x][next.y]||a1[next.x][next.y]==||a2[next.x][next.y]==)
continue;
vis[next.x][next.y]=;
q.push((node){next.x,next.y,next.z});
}
}
if(flag)
puts("YES");
else
puts("NO"); }