# Tunnels

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description
Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his eyes, the city is a grid. He can’t enter a grid with a barrier. In one minute, he can move into an adjacent grid with no barrier. Bob is full of curiosity and he wants to visit all of the secret tunnels beneath the city. To travel in a tunnel, he has to walk to the entrance of the tunnel and go out from the exit after a fabulous visit. He can choose where he starts and he will travel each of the tunnels once and only once. Now he wants to know, how long it will take him to visit all the tunnels (excluding the time when he is in the tunnels).

Input
The input contains mutiple testcases. Please process till EOF.
For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.
The map of the city is given in the next N lines. Each line contains exactly N characters. Barrier is represented by “#” and empty grid is represented by “.”.
Then M lines follow. Each line consists of four integers x1, y1, x2, y2, indicating there is a tunnel with entrence in (x1, y1) and exit in (x2, y2). It’s guaranteed that (x1, y1) and (x2, y2) in the map are both empty grid.

Output
For each case, output a integer indicating the minimal time Bob will use in total to walk between tunnels.
If it is impossible for Bob to visit all the tunnels, output -1.

Sample Input
5 4 ....# ...#. ..... ..... ..... 2 3 1 4 1 2 3 5 2 3 3 1 5 4 2 1

Sample Output
7

Source

dp[i][j]=dp[i-(1<<(j-1)][k] +dis(k,j)  dis(k,j)表示第k条通道的终点到第j条通道的起点距离；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<bitset>
#include<time.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-8
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=10+10,M=1e6+10,inf=1e9+7,MOD=1e9+7;
const LL INF=1e18+10,mod=1e9+7;

int n,m,vis[N][N];
char a[N][N];
struct is
{
int s,t,e,d;
} q[N];
int check(int x,int y)
{
if(x<=0||x>n||y<=0||y>n)return 0;
return 1;
}
int dis[N][N][N][N];
int xx[5]= {1,0,-1,0};
int yy[5]= {0,1,0,-1};
void bfs(int s,int t)
{
queue<pair<int,int> >q;
q.push(make_pair(s,t));
memset(vis,0,sizeof(vis));
vis[s][t]=1;dis[s][t][s][t]=0;
while(!q.empty())
{
pair<int,int> p=q.front();
q.pop();
for(int i=0; i<4; i++)
{
int x=p.first+xx[i];
int y=p.second+yy[i];
if(check(x,y)&&!vis[x][y]&&a[x][y]=='.')
{
vis[x][y]=1;
dis[s][t][x][y]=dis[s][t][p.first][p.second]+1;
q.push(make_pair(x,y));
}
}
}
}
int dp[50000][N];
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(dis,-1,sizeof(dis));
for(int i=1; i<=n; i++)
scanf("%s",a[i]+1);
for(int i=1; i<=m; i++)
scanf("%d%d%d%d",&q[i].s,&q[i].t,&q[i].e,&q[i].d);
for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)bfs(i,j);
memset(dp,-1,sizeof(dp));
for(int i=1;i<=(1<<m)-1;i++)
{
for(int j=1;j<=m;j++)
{
if((1<<(j-1))&i)
{
int now=i-(1<<(j-1));
if(!now)
{
dp[i][j]=0;
continue;
}
for(int k=1;k<=m;k++)
{
if(now&(1<<(k-1)))
{
if(dp[now][k]!=-1&&dis[q[k].e][q[k].d][q[j].s][q[j].t]!=-1)
{
int temp=dp[i][j];
dp[i][j]=dp[now][k]+dis[q[k].e][q[k].d][q[j].s][q[j].t];
if(temp!=-1)dp[i][j]=min(dp[i][j],temp);
}
}
}
}
}
}
int ans=inf;
for(int i=1;i<=m;i++)
if(dp[(1<<m)-1][i]!=-1)ans=min(ans,dp[(1<<m)-1][i]);
if(ans!=inf)printf("%d\n",ans);
else printf("-1\n");
}
return 0;
}

posted @ 2017-10-11 20:48  jhz033  阅读(214)  评论(0编辑  收藏  举报