题意：给你一个图，并告诉你只能走那些点，问你从0走完这S个点在回去的最短路是多少
思路：首先预处理出从这S+1个点（0和S个点）两两到达的距离，之后就是一发裸的状压就好了
上代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000+10;
const int INF = 9999999;
struct node
{
int to,nex,val;
}edg[maxn<<1];
int n,m,cnt;
int head[maxn],dis[maxn],vis[maxn];
int DIS[12][12];
int rode[maxn],dp[1<<11][11];
void init()
{
cnt = 0 ;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
edg[cnt].to = v;
edg[cnt].val = w;
edg[cnt].nex = head[u];
head[u] = cnt++;
}
int SPFA(int s,int e)
{
for(int i = 0 ; i < n ; i++)
{
dis[i] = INF;
vis[i] = 0;
}
queue<int>que;
while(!que.empty()) que.pop();
que.push(s);
dis[s] = 0;
while(!que.empty())
{
int u = que.front();
que.pop();
vis[u] = 0;
for(int i = head[u] ; i != -1 ; i = edg[i].nex)
{
int v = edg[i].to;
int w = edg[i].val;
if(dis[v] > dis[u] + w)
{
dis[v] = dis[u] + w;
if(!vis[v]){
vis[v] = 1;
que.push(v);
}
}
}
}
return dis[e];
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
memset(DIS,0,sizeof(DIS));
init();
int a,b,c;
for(int i = 0 ; i < m ; i++)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
for(int i = 0 ; i < n ; i++)
{
for(int j = head[i] ; j!= -1 ; j = edg[j].nex)
{
int v = edg[j].to;
int w = edg[j].val;
}
}
int te;
scanf("%d",&te);
for(int i = 1 ; i <= te ; i ++)
{
scanf("%d",&rode[i]);
}
for(int i = 1 ; i <= te ; i++)
{
DIS[0][i] = DIS[i][0] = SPFA(0,rode[i]);
}
for(int i = 1 ; i <= te ; i ++)
{
for(int j = i + 1 ; j <= te ; j++)
{
DIS[i][j] = DIS[j][i] = SPFA(rode[i],rode[j]);
}
}
/* for(int i = 0 ; i <= te ; i++)
{
for(int j = 0 ; j <= te ; j ++)
{
printf("%d ",DIS[i][j]);
}
puts("");
}*/
te++;
for(int i = 0 ; i < (1<<te) ; i++)
{
for(int j = 0 ; j < te ; j ++) dp[i][j] = INF;
}
dp[0][0] = 0;
for(int state = 0 ; state < (1<<te) ; state++)
{
for(int i = 0 ; i < te ; i ++)
{
if(dp[state][i] == INF) continue;
for(int j = 0 ; j < te ; j++)
{
if((1<<j)&state) continue;
if(DIS[i][j] == INF) continue;
dp[(1<<j)|state][j] = min(dp[(1<<j)|state][j] , dp[state][i] + DIS[i][j]);
}
}
}
int ans = INF;
for(int i = 0 ;i < te ; i++)
{
// printf("%d %d\n",dp[(1<<te)-1][i] , DIS[i][0]);
ans = min(ans,dp[(1<<te)-1][i] + DIS[i][0]);
}
printf("%d\n",ans);
}
return 0;
}