hdu 4568 spfa 最短路演算法+旅行商問題

http://acm.hdu.edu.cn/showproblem.php?pid=4568

　　One day, a hunter named James went to a mysterious area to find the treasures. James wanted to research the area and brought all treasures that he could.
　　The area can be represented as a N*M rectangle. Any points of the rectangle is a number means the cost of research it,-1 means James can't cross it, James can start at any place out of the rectangle, and explore point next by next. He will move in the rectangle and bring out all treasures he can take. Of course, he will end at any border to go out of rectangle(James will research every point at anytime he cross because he can't remember whether the point are researched or not).
　　Now give you a map of the area, you must calculate the least cost that James bring out all treasures he can take(one point up to only one treasure).Also, if nothing James can get, please output 0.

/**
hdu 4568 spfa 最短路演算法+旅行商問題
題目大意：給定一個n*m的棋盤，每一個格子有一個值，代表經過這個格子的花費，給出sum個寶藏點的座標，求從棋盤的任意一個邊進入棋盤，經過所有的寶藏點後在走出
          棋盤所需要的最小花費
解題思路：spfa處理處任意兩個寶藏點之間的最短距離（最小花費）和每個寶藏點和邊界的最短距離。然後狀態壓縮：dp[s][i]表示經過寶藏點的狀態為s並且結尾點為i的
          最小花費
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
const int maxn=205;
int n,m;
struct note
{
    int x,y;
} point[15];

int dx[4][2]= {1,0,0,1,-1,0,0,-1};
int dis[maxn][maxn],a[maxn][maxn],dis_border[25],length[20][20];
bool vis[maxn][maxn];
int dp[1<<15][15];

void spfa(int s)
{
    for(int i=0;i<n;i++)
         for(int j=0;j<m;j++)
            dis[i][j]=0x3f3f3f3f;
    memset(vis,0,sizeof(vis));
    queue<pair<int,int> >q;
    q.push(make_pair(point[s].x,point[s].y));
    vis[point[s].x][point[s].y]=1;
    dis[point[s].x][point[s].y]=0;
    while(!q.empty())
    {
        int x=q.front().first;
        int y=q.front().second;
        q.pop();
        vis[x][y]=0;
        if(x==0||x==n-1||y==0||y==m-1)
            dis_border[s]=min(dis_border[s],dis[x][y]);
        for(int i=0; i<4; i++)
        {
            int xx=x+dx[i][0];
            int yy=y+dx[i][1];
            if(xx>=0&&xx<n&&yy>=0&&yy<m&&a[xx][yy]!=-1)
            {
                if(dis[xx][yy]>dis[x][y]+a[xx][yy])
                {
                    dis[xx][yy]=dis[x][y]+a[xx][yy];
                    if(vis[xx][yy]==0)
                    {
                        vis[xx][yy]=1;
                        q.push(make_pair(xx,yy));
                    }
                }
            }
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        int k;
        scanf("%d\n",&k);
        for(int i=0;i<k;i++)
        {
            scanf("%d%d",&point[i].x,&point[i].y);
        }
        //===預處理===
        for(int i=0;i<k;i++)
        {
            dis_border[i]=0x3f3f3f3f;
            for(int j=0;j<k;j++)
            {
                if(i==j)
                    length[i][j]=0;
                else
                    length[i][j]=0x3f3f3f3f;
            }
        }
        for(int i=0;i<(1<<k);i++)
        {
            for(int j=0;j<k;j++)
            {
                dp[i][j]=0x3f3f3f3f;
            }
        }
        //===求有寶藏的點之間和每一個寶藏點和邊界的最短距離===
        for(int i=0; i<k; i++)
        {
            spfa(i);
            for(int j=0; j<k; j++)
            {
                if(j==i)continue;
                length[i][j]=min(dis[point[j].x][point[j].y],length[i][j]);
            }
            dp[1<<i][i]=dis_border[i]+a[point[i].x][point[i].y];
        }
        ///===求最優路徑===
        for(int s=0;s<(1<<k);s++)
        {
            for(int i=0;i<k;i++)
            {
                if(s&(1<<i)==0)continue;
                if(dp[s][i]==0x3f3f3f3f)continue;
                for(int j=0;j<k;j++)
                {
                    if(s&(1<<j)==1)continue;
                    dp[s|(1<<j)][j]=min(dp[s|(1<<j)][j],dp[s][i]+length[i][j]);
                }
            }
        }
        ///===還要回到邊界==
        int ans=0x3f3f3f3f;
        for(int i=0;i<k;i++)
        {
            ans=min(ans,dp[(1<<k)-1][i]+dis_border[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}