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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 2014西安全國邀請賽

題意：一個圖，#不可達，m條單向通道，點間花費1時間，通道起點到終點不花費時間，求最少花費時間；

思路：狀態壓縮dp，dp[i][j] i表示已經走過哪些通道，j表示最後的走的那條通道是哪條。

　　　dp[i][j]=dp[i-(1<<(j-1)][k] +dis(k,j) dis(k,j)表示第k條通道的終點到第j條通道的起點距離；

#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;
}

hdu 4856 Tunnels 狀態壓縮dp