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Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/gym/100187/problem/E
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
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=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** int a1[510][510]; int a2[510][510]; string s; struct node { int x,y,z; }; int dx[4]={1,-1,0,0}; int dy[4]={0,0,1,-1}; int vis[510][510]; int main() { int n=read(),m=read(); for(int i=0;i<n;i++) { cin>>s; for(int j=0;j<m;j++) { if(s[j]==‘.‘) a1[i][j]=1; else a1[i][j]=0; } } for(int i=0;i<n;i++) { cin>>s; for(int j=0;j<m;j++) { if(s[j]==‘.‘) a2[i][j]=1; else a2[i][j]=0; } } queue<node> q; q.push((node){0,0,0}); int flag=0; int ans1=inf; memset(vis,0,sizeof(vis)); vis[0][0]=1; while(!q.empty()) { node now=q.front(); q.pop(); if(now.x==n-1&&now.y==m-1&&ans1>now.z) { ans1=now.z; continue; } for(int i=0;i<4;i++) { node next; next.x=now.x+dx[i]; next.y=now.y+dy[i]; next.z=now.z; if(next.x<0||next.x>=n) continue; if(next.y<0||next.y>=m) continue; if(vis[next.x][next.y]||a1[next.x][next.y]==0) continue; vis[next.x][next.y]=1; q.push((node){next.x,next.y,next.z+1}); } } while(!q.empty()) q.pop(); q.push((node){0,0,0}); int ans2=inf; memset(vis,0,sizeof(vis)); vis[0][0]=1; while(!q.empty()) { node now=q.front(); q.pop(); if(now.x==n-1&&now.y==m-1&&ans2>now.z) { ans2=now.z; continue; } for(int i=0;i<4;i++) { node next; next.x=now.x+dx[i]; next.y=now.y+dy[i]; next.z=now.z; if(next.x<0||next.x>=n) continue; if(next.y<0||next.y>=m) continue; if(vis[next.x][next.y]||a2[next.x][next.y]==0) continue; vis[next.x][next.y]=1; q.push((node){next.x,next.y,next.z+1}); } } if(ans1!=ans2) { puts("NO"); return 0; } while(!q.empty()) q.pop(); q.push((node){0,0,0}); memset(vis,0,sizeof(vis)); vis[0][0]=1; while(!q.empty()) { if(flag) break; node now=q.front(); q.pop(); if(now.x==n-1&&now.y==m-1&&now.z==ans1) flag=1; if(flag) break; for(int i=0;i<4;i++) { node next; next.x=now.x+dx[i]; next.y=now.y+dy[i]; next.z=now.z+1; if(next.x<0||next.x>=n) continue; if(next.y<0||next.y>=m) continue; if(vis[next.x][next.y]||a1[next.x][next.y]==0||a2[next.x][next.y]==0) continue; vis[next.x][next.y]=1; q.push((node){next.x,next.y,next.z}); } } if(flag) puts("YES"); else puts("NO"); }
Codeforces Gym 100187E E. Two Labyrinths bfs
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原文地址：http://www.cnblogs.com/qscqesze/p/4657681.html
