Hdu 4568 Hunter【spfa最短路 tsp状态压缩】

Description

　　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.

 

Input

　　The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case begins with a line containing 2 integers N M , (1<=N,M<=200), that represents the rectangle. Each of the following N lines contains M numbers(0~9),represent the cost of each point. Next is K(1<=K<=13),and next K lines, each line contains 2 integers x y means the position of the treasures, x means row and start from 0, y means column start from 0 too.

 

Output

　　For each test case, you should output only a number means the minimum cost.

 

Sample Input

     
     
     
     

      
      
      
      2
3 3
3 2 3
5 4 3
1 4 2
1
1 1
3 3
3 2 3
5 4 3
1 4 2
2
1 1
2 2 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      8
11 
     
     
     
     

 

这题太麻烦了==尤其是tsp没好好学的情况下  其中dp[i][j]: 从j出发，经过状态 i 上所有点的最短距离

    #include<cstdio>
    #include<cstring>
    #include<queue>
    #include<algorithm>
    using namespace std;
    #define MAX 205
    #define inf 99999999
    int n,m;
    struct pos{int x,y;}tr_pos[MAX];
    int map[MAX * MAX],tr,val[14],dis[MAX * MAX];
    int dir[4][2]={0,1,0,-1,-1,0,1,0};
    int g[20][20],dp[1<<14][14];
    int get_id(int x,int y){
        if(x>=0 && x<n && y>=0 && y<m)return x*m+y;
        return n*m;
    }
    void spfa(int x){
        for (int i = 0; i <= n*m; i++)
            dis[i] = inf;//dis数组是对应当前财宝所在位置距外部的最短距离，需要每次都初始化
        dis[x] = (map[x]==-1?inf:map[x]);
        bool inq[205*205];
        memset(inq,0,sizeof(inq));
        queue<int>Q;
        while(!Q.empty())Q.pop();
        Q.push(x);
        inq[x] = 1;
        int now;
        while(!Q.empty()){
            now = Q.front();
            Q.pop();
            inq[now] = 0;
            if(now == n*m)continue;
            int r = now/m;
            int c = now%m;
            int nr,nc,nid;
            for (int i = 0; i < 4; i++)
            {
                nr = r + dir[i][0];
                nc = c + dir[i][1];
                nid = get_id(nr,nc);
                if(map[nid] == -1)continue;
                //更新 财宝点到边界的最短距离。
                if(dis[nid] > dis[now] + map[nid]){
                    dis[nid] = dis[now] + map[nid];
                    if(!inq[nid])   Q.push(nid);
                    inq[nid] = 1;
                }
            }
        }
    }
    int done(){
        for (int i = 0; i < tr; i++)
        {
            spfa(get_id(tr_pos[i].x,tr_pos[i].y));
            for (int j = 0; j < tr; j++)
            {
                int jid = get_id(tr_pos[j].x,tr_pos[j].y);
                g[i][j] = dis[jid];
                if(g[i][j] == inf) return 0;
            }
            g[i][tr] = dis[n*m];//将n*m点看做一个花费为0的财宝点。
        }
        return 1;
    }
    int solve()
    {
        if(tr == 0) return 0;
        int full = 1<<tr;
        for(int i = 0;i<full;i++)
            for(int j=0;j<tr;j++)
                dp[i][j] = inf;
        for (int i = 0; i < tr; i++)
            dp[1<<i][i] = g[i][tr];
        for(int i = 1; i < full; ++i){   //当前状态为 i 状态（用二进制表示经过了哪些财宝点）
        for(int j = 0; j < tr; ++j){
            if(i & (1<<j)){               //保证第j个财宝点在 i 状态中。
        for(int k = 0; k < tr; ++k){
            if( i & (1<<k) && j != k){    //第k个也要在，而且i！= j
                if(dp[i^(1<<k)][j] != inf)
                dp[i][k] = min(dp[i][k], dp[i^(1<<k) ][j] + g[j][k] - val[j]);
            }
            }
        }
        }
        }
        int res = inf;
        for (int i = 0; i < tr; i++)
        {
            res = min(res,dp[full-1][i] + g[i][tr]-val[i]);
            //full-1表示所有财宝点都包含的状态,g[][]表示两财宝点之间的最短距离，因为i的val记了两遍，故减去
        }
        return res;
    }
    int main()
    {
        int cas,i,j;
        scanf("%d",&cas);
        while(cas--){
            scanf("%d%d",&n,&m);
            for(i = 0;i < n;i++)
            for(j = 0;j < m;j++)
                scanf("%d",&map[i*m+j]);
            scanf("%d",&tr);
            for(i = 0;i < tr;i++){
                scanf("%d%d",&tr_pos[i].x,&tr_pos[i].y);
                int id = get_id(tr_pos[i].x,tr_pos[i].y);
                val[i] = map[id];
            }
            map[n*m] = 0;//从外部任意位置进入。

            if(done())printf("%d\n",solve());
            else printf("-1\n");
        }
        return 0;
    }