[状压dp,TSP问题] 2020牛客国庆集训派对day2-B CHEAP DELIVERIES

题目：https://ac.nowcoder.com/acm/contest/7818/B

经典TSP问题，推荐看这篇文章：https://www.cnblogs.com/smashfun/p/11432110.html

模版：

for (int i = 1; i <= n; i++)  
    dp[1<<(i - 1)][i] = 0;  
int ans = inf;  
for (int i = 0; i < (1<<n); i++) //状态的个数，从n个0到n个1 
    for (int j = 1; j <= n; j++) 
    {  
        if (i&(1<<(j - 1))) //必须是访问过的点j才可以，访问过的才知道最优子结构
        {  
            for (int k = 1; k <= n; k++) 
                if (i&(1<<(k - 1)) != 0 && g[k][j] != -1) //必须是访问过的k点且边存在
                    dp[i][j] = min(dp[i^(1<<(j - 1))][k] + g[k][j], dp[i][j]);  
        }  
        if (i == (1<<n) - 1)  
            ans = min(ans, dp[i][j]);  
    }  
if (ans == inf) 
    return -1;  
return ans;

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
//typedef __int128_t LL;
typedef double db;
#define rep(a,b,c) for(ll a=b;a<=c;a++)
#define per(a,b,c) for(ll a=b;a>=c;a--)
#define go(a,b) for(ll a=head[b];a;a=e[a].to)
#define endl '\n'
#define V vector
#define pb push_back
#define mp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
const ll amn=1e5+5,mod=1e9+7,inf=0x3f3f3f3f;
ll head[amn],etot;
struct eg{
    ll to,v;
    db w;
    eg(){}
    eg(ll to,ll v,db w):to(to),v(v),w(w){}
}e[amn];
void einit(){
    etot=0;
    mem(head,0);
}
void add(ll u,ll v,db w){
    e[++etot]=eg(head[u],v,w);
    head[u]=etot;
}
void sovle();
int main(){
    ios::sync_with_stdio(0);
    ll T=1;
    //cin>>T;
    while(T--){
        sovle();
    }
}
ll n,m,k;
ll s,t;
struct Node{
    int id; db d;
    Node() {}
    Node(int id, db d):id(id),d(d){}
    bool operator < (const Node &A) const {
        return d > A.d;
    }
};
ll vis[amn];
db dis[amn];
void dij(int st){
    for(int i=1; i<=n; i++){
        vis[i] = 0;
        dis[i] = inf;
    }

    dis[st] = 0;
    priority_queue <Node> Q;
    while(Q.size())Q.pop();
    Q.push(Node(st, 0));
    Node nd;

    while(!Q.empty()){
        nd = Q.top(); Q.pop();
        if(vis[nd.id]) continue;
        vis[nd.id] = true;
        go(i,nd.id){
            int j = e[i].v;
            db k = e[i].w;
            if(nd.d + k < dis[j] && !vis[j]){
                dis[j] = nd.d + k;
                Q.push(Node(j, dis[j]));
            }
        }
    }
}
ll kl[55][5];
ll d[55][55];
ll dp[1<<20][25];
void sovle(){
    ll ans=inf;
    mem(head,0);
    mem(d,0x3f);
    cin>>n>>m>>k;
    rep(i,1,m){
        ll u,v,w;
        cin>>u>>v>>w;
        add(u,v,w);
        add(v,u,w);
    }
    rep(i,1,k){
       rep(j,0,1){
           cin>>kl[i][j];
       }
    }
    rep(i,1,k){
        dij(kl[i][0]);
        rep(j,1,k){
            d[i][j]=dis[kl[j][1]];
        }
    }
    mem(dp,0x3f);
    rep(i,0,k-1){
        dp[1<<(i)][i+1]=d[i+1][i+1];
    }
    rep(i,0,(1<<k)-1){
        rep(j,0,k-1){
            if(i&(1<<(j))){
                rep(v,0,k-1){
                    if((i&(1<<(v)))!=0&&d[j+1][v+1]!=inf){
                        dp[i][j+1]=min(dp[i][j+1],dp[i^(1<<(j))][v+1]+d[j+1][v+1]+d[j+1][j+1]);
                    }
                }
                if(i==(1<<(k))-1){
                    ans=min(ans,dp[i][j+1]);
                }
            }
        }
    }
    if(ans==inf)ans=-1;
    cout<<ans<<endl;
}
/*
5 5 3 
1 2 1 
2 3 2 
3 4 3 
4 5 4 
5 2 4 
2 3 
1 2 
5 3
*/