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;
}