Light OJ 1316 A Wedding Party 最短路+状态压缩DP

题目来源：Light OJ 1316 1316 - A Wedding Party

题意：和HDU 4284 差点儿相同 有一些商店 从起点到终点在走过尽量多商店的情况下求最短路

思路：首先预处理每两点之前的最短路 然后仅仅考虑那些商店 个数小于15嘛 就是TSP问题 状态压缩DP搞一下 状态压缩姿势不正确 有必要加强

#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;
const int maxn = 510;
const int maxm = 16;
const int INF = 999999999;
struct edge
{
    int u, v, w;
	edge(){}
	edge(int u, int v, int w): u(u), v(v), w(w) {}
};

struct HeapNode
{
    int u, dis;
    HeapNode(){};
    HeapNode(int u, int dis): u(u), dis(dis){};
    bool operator < (const HeapNode& rhs)const
    {
        return dis > rhs.dis;
    }
};
vector <edge> G[maxn];
int d[maxn][maxn];
int dp[1<<maxm][maxm];
bool vis[maxn];
int n, m, t;
int a[maxm];
void Dijkstra(int s)
{
    for(int i = 0; i <= n; i++)
   		d[s][i] = INF;
	d[s][s] = 0;
    memset(vis, false, sizeof(vis));
    priority_queue <HeapNode> Q;
    Q.push(HeapNode(s, 0));
    while(!Q.empty())
    {
        HeapNode x = Q.top();
        Q.pop();
        int u = x.u;
        if(vis[u])
            continue;
        vis[u] = true;
        for(int i = 0; i < G[u].size(); i++)
        {
            edge e = G[u][i];
            int v = e.v;
            if(d[s][v] > x.dis + e.w)
            {
                d[s][v] = x.dis + e.w;
                Q.push(HeapNode(v, d[s][v]));
            }
        }
    }
}
int get(int x)
{
	int ans = 0;
	while(x)
	{
		if(x&1)
			ans++;
		x >>= 1;
	}
	return ans;
}
int main()
{
	int T;
    int cas = 0;
    scanf("%d", &T);
    while(T--)
    {
    	scanf("%d %d %d", &n, &m, &t);
        for(int i = 0; i <= n; i++)
            G[i].clear();
        for(int i = 0; i < t; i++)
        {
        	int x;
        	scanf("%d", &x);
        	a[i] = x;
        }
        for(int i = 0; i < m; i++)
        {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
			G[u].push_back(edge(u, v, w));
        }
        for(int i = 0; i < n; i++)
 	 		Dijkstra(i);
	  	for(int s = 0; s < (1<<t); s++)
   		{
			for(int i = 0; i < t; i++)
 			{
        		dp[s][i] = INF;
				if(!(s&(1<<i)))
        			continue;
        		if(s == (1<<i))
       			{
       				//if(s == 2 && i == 1)
       				//	printf("%d\n", d[0][a[i]]);
 					dp[s][i] = d[0][a[i]];
 					continue;
       			}
       			for(int j = 0; j < t; j++)
       			{
       				if((s&(1<<j)) && (i != j))
       				{
       					if(dp[s^(1<<i)][j] == INF)
       						continue;
    					if(d[a[j]][a[i]] == INF)
							continue;
						//if(s == 3 && i == 0)
						//	printf("%d %d %d %d\n", dp[s^(1<<i)][j], d[a[j]][a[i]], j, dp[2][1]);
       					dp[s][i] = min(dp[s^(1<<i)][j] + d[a[j]][a[i]], dp[s][i]);
       				}
       			}

        	}
        }
        //printf("222*%d\n", dp[3][0]);
        int x;
        int ans = INF, sum = 0;
        for(int s = 1; s < (1<<t); s++)
        {

        	for(int i = 0; i < t; i++)
        	{
				//if(s == (1<<i))
        		//	printf("***%d %d %d\n", dp[s][i], i, s);
				//printf("**%d %d %d %d\n", dp[s][i], s, i, dp[2][i]);
				if(dp[s][i] == INF || d[a[i]][n-1] == INF)
        			continue;
        		int temp = get(s);
        		if(sum < temp || sum == temp && ans > dp[s][i]+d[a[i]][n-1])
        		{

        			sum = temp;
        			ans = dp[s][i]+d[a[i]][n-1];
        			x = s;
        		}
        	}
        }
        if(sum == 0)
        {
            printf("Case %d: Impossible\n", ++cas);
            continue;
        }
		printf("Case %d: %d %d\n", ++cas, sum, ans);
    }
    return 0;
}