HDU 1535 Invitation Cards 2次Dijkstra来回最短路

题目来源：HDU 1535 Invitation Cards

题意：从1派学生到2-n这n-1个点 求去并且回来的最短路 就是1到各点的最短路之和和各点到1的最短路之和 给的是有向图

思路：对于1到各个点的最短路直接Dijkstra求出无压力 然后各个点到1的最短路可以反向建图后再求一次从1到各个点的最短路

对于很多点到一个点的情况可以考虑反向建图 转变成单源最短路

#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 1000010;
struct edge
{
	int u, v, w;
};
struct HeapNode
{
	int u, d;
	bool operator < (const HeapNode& rhs)const
	{
		return d > rhs.d;
	}
};
vector <edge> edges;
vector <edge> G[maxn];
int dis[maxn];
bool vis[maxn];
int n, m;
void Dijkstra()
{
	//for(int i = 0; i <= n; i++)
	//	dis[i] = 9999999999;
	memset(dis, 0x7f, sizeof(dis));
	dis[1] = 0;
	memset(vis, false, sizeof(vis));
	priority_queue <HeapNode> Q;
	Q.push((HeapNode){1, 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(dis[v] > x.d + e.w)
			{
				dis[v] = x.d + e.w;
				Q.push((HeapNode){v, dis[v]});	
			}
		}
	}
}
int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d", &n, &m);
		edges.clear();
		for(int i = 0; i < m; i++)
		{
			int u, v, w;
			scanf("%d %d %d", &u, &v, &w);
			edges.push_back((edge){u, v, w});
			
		}
		int ans = 0;
		for(int i = 0; i <= n; i++)
			G[i].clear();
		for(int i = 0; i < m; i++)
		{
			edge e = edges[i];
			G[e.u].push_back((edge){e.u, e.v, e.w});
		}
		Dijkstra();
		for(int i = 2; i <= n; i++)
			ans += dis[i];
			
		for(int i = 0; i <= n; i++)
			G[i].clear();
		for(int i = 0; i < m; i++)
		{
			edge e = edges[i];
			G[e.v].push_back((edge){e.v, e.u, e.w});
		}
		Dijkstra();
		for(int i = 2; i <= n; i++)
			ans += dis[i];
		printf("%d\n", ans);
	}
	return 0;
}