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E. Two Labyrinths
time limit per test
memory limit per test
input
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.
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
n and m (1 ≤ n, m ≤ 500) are written — the height and the width of the labyrinths.
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.
n
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
3 5
.....
.#.#.
.....
.....
#.#.#
.....
output
NO
input
3 5
.....
.#.##
.....
.....
##.#.
.....
output
YES
大意：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：跑三发 bfs，第一次找一图的最短路，第二次找二图的最短路，第三次核查两图的最短路是否相同。刚开始以为 dfs 能闯过去，交了一发果断 tle
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
int n,m;
char mp[2][510][510];
bool vis[2][510][510];
bool mark[510][510];
int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};
int ans[2];
struct node
{
int x,y,step;
};
bool judge(int x,int y)
{
if(x<0||y<0||x>=n||y>=m)
return 0;
return 1;
}
queue<node> Q;
void bfs(int id)
{
while(!Q.empty()) Q.pop();
node now,next;
now.x=0,now.y=0,now.step=0;
vis[id][0][0]=1;
Q.push(now);
while(!Q.empty())
{
now=Q.front();
Q.pop();
if(now.x==n-1&&now.y==m-1)
{
ans[id]=now.step;
return ;
}
for(int i=0;i<4;i++)
{
next.x=now.x+dx[i];
next.y=now.y+dy[i];
if(judge(next.x,next.y)&&mp[id][next.x][next.y]=='.'&&vis[id][next.x][next.y]==0)
{
next.step=now.step+1;
vis[id][next.x][next.y]=1;
Q.push(next);
}
}
}
}
bool flag;
void find()
{
while(!Q.empty()) Q.pop();
node now,next;
now.x=0,now.y=0,now.step=0;
mark[0][0]=1;
Q.push(now);
while(!Q.empty())
{
now=Q.front();
Q.pop();
if(now.x==n-1&&now.y==m-1&&now.step==ans[0]) // 这里要加判断步数才能 ac
{
flag=1;
return ;
}
for(int i=0;i<4;i++)
{
next.x=now.x+dx[i];
next.y=now.y+dy[i];
if(judge(next.x,next.y)&&mark[next.x][next.y]==0&&mp[0][next.x][next.y]=='.'&&mp[1][next.x][next.y]=='.')
{
next.step=now.step+1;
mark[next.x][next.y]=1;
Q.push(next);
}
}
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
for(int i=0;i<n;i++)
scanf("%s",mp[0][i]);
for(int i=0;i<n;i++)
scanf("%s",mp[1][i]);
ans[0]=INF,ans[1]=INF,ans[2]=INF;
memset(vis,0,sizeof(vis));
bfs(0); bfs(1);
if(ans[0]!=ans[1])
{
puts("NO");
continue;
}
memset(mark,0,sizeof(mark));
flag=0;
find();
// printf("%d %d %d\n",ans[0],ans[1],ans[2]);
if(flag)
puts("YES");
else
puts("NO");
}
return 0;
}
