poj--1383--Labyrinth(树的直径)
Time Limit: 2000MS | Memory Limit: 32768K | |
Total Submissions: 4062 | Accepted: 1529 |
Description
found that two of the hooks must be connected by a rope that runs through the hooks in every block on the path between the connected ones. When the rope is fastened, a secret door opens. The problem is that we do not know which hooks to connect. That means
also that the neccessary length of the rope is unknown. Your task is to determine the maximum length of the rope we could need for a given labyrinth.
Input
each containing C characters. These characters specify the labyrinth. Each of them is either a hash mark (#) or a period (.). Hash marks represent rocks, periods are free blocks. It is possible to walk between neighbouring blocks only, where neighbouring blocks
are blocks sharing a common side. We cannot walk diagonally and we cannot step out of the labyrinth.
The labyrinth is designed in such a way that there is exactly one path between any two free blocks. Consequently, if we find the proper hooks to connect, it is easy to find the right path connecting them.
Output
Sample Input
2
3 3
###
#.#
###
7 6
#######
#.#.###
#.#.###
#.#.#.#
#.....#
#######
Sample Output
Maximum rope length is 0.
Maximum rope length is 8.
Hint
If you use recursion, maybe stack overflow. and now C++/c 's stack size is larger than G++/gcc
Source
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
struct node
{
int x,y,step;
}temp,p;
int vis[1010][1010],sx,sy,ans,m,n;
char map[1010][1010];
void init()
{
memset(map,'\0',sizeof(map));
memset(vis,0,sizeof(vis));
ans=0;
sx=sy=0;
}
void getmap()
{
int flag=0;
for(int i=0;i<m;i++)
{
scanf("%s",map[i]);
for(int j=0;j<n&&!flag;j++)
{
if(map[i][j]=='.')
{
sx=i;
sy=j;
flag=1;
}
}
}
}
int judge(node s1)
{
if(s1.x<0||s1.x>=m||s1.y<0||s1.y>=n)
return 1;
if(map[s1.x][s1.y]=='#'||vis[s1.x][s1.y])
return 1;
return 0;
}
void bfs(int x,int y)
{
memset(vis,0,sizeof(vis));
queue<node>q;
p.x=sx;
p.y=sy;
p.step=0;
q.push(p);
vis[sx][sy]=1;
while(!q.empty())
{
p=q.front();
q.pop();
for(int i=0;i<4;i++)
{
temp.x=p.x+dx[i];
temp.y=p.y+dy[i];
if(judge(temp)) continue;
temp.step=p.step+1;
if(temp.step>ans)
{
ans=temp.step;
sx=temp.x;
sy=temp.y;
}
vis[temp.x][temp.y]=1;
q.push(temp);
}
}
}
void solve()
{
bfs(sx,sy);
bfs(sx,sy);
printf("Maximum rope length is %d.\n",ans);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
getmap();
solve();
}
return 0;
}
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