HDU 1535 Invitation Cards(最短路)

题目读了半天，发现是从1到 2 ~n个点再从 2 ~n个点回到1的最短路。。

所以正向反向分别建图，跑最短路就好了。

#include <algorithm>
#include <queue>
#include <cstring>
#include <iostream>

using namespace std;
typedef pair<int,int> pa;
#define N 1000005
#define inf 0x5f5f5f5f
class op{
public:
    bool operator()(const pa&a,const pa&b){
        return a.first>b.first;
    }
};
class graphic{
public:
    void init(){
        cnt = 0;
        memset(head,-1, sizeof(head));
    }
    void add(int u,int v,int d){
        edge[cnt].to = v;
        edge[cnt].dis = d;
        edge[cnt].next = head[u];
        head[u] = cnt++;
    }
    long long deal(int n){
        dijkstra(1);
        long long ans = 0;
        for(int i = 2 ; i <= n ; ++i){
            ans+=dis[i];
        }
        return ans;
    }
private:
    struct Edge{
        int to,next,dis;
    }edge[N];
    int head[N],cnt;
    int dis[N];
    priority_queue <pa,vector<pa>,op>ss;
    void dijkstra(int beg){
        memset(dis,inf, sizeof(dis));
        dis[beg] = 0;
        while (!ss.empty())ss.pop();
        ss.emplace(make_pair(dis[0],beg));
        while (!ss.empty()){
            int u = ss.top().second;
            ss.pop();
            for(int i = head[u] ; ~i ; i = edge[i].next){
                if(dis[edge[i].to] > dis[u] + edge[i].dis){
                    dis[edge[i].to] = dis[u] + edge[i].dis;
                    ss.emplace(make_pair(dis[edge[i].to],edge[i].to));
                }
            }
        }
    }
};
graphic from,to;
int main() {
    int T;
    cin>>T;
    int n,m,u,v,d;
    while (T--){
        cin>>n>>m;
        from.init();
        to.init();
        for(int i = 0 ; i < m ; ++i){
            scanf("%d %d %d",&u,&v,&d);
            from.add(u,v,d);
            to.add(v,u,d);
        }
        long long ans = from.deal(n)+to.deal(n);
        cout<<ans<<"
";
    }
}