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;
}