hdu 1513 Invitation Cards【spfa翻转边】

题目链接：http://acm.acmcoder.com/showproblem.php?pid=1535

题意：有向图，求源点到各个点最短路径和+各个点到源点最短路径和。

spfa求单源最短路径，求各个点到源点最短路径翻转边。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

const int MAXN = 2000010;
const int INF = 1e9+10;

struct
{
    int a, b, c;
}Map[MAXN];

int t;
int n, m, a, b, c;

struct Edge
{
    int v;
    int cost;
    Edge(int _v = 0, int _cost = 0)
    {
        v = _v;
        cost = _cost;
    }
};
vector<Edge> E[MAXN];

void addedge(int u, int v, int cost)
{
    E[u].push_back(Edge(v, cost));
}

bool vis[MAXN];
int cnt[MAXN];
int dist[MAXN];

void SPFA(int start, int n)
{
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; i++) dist[i] = INF;
    vis[start] = true;
    dist[start] = 0;
    queue<int> que;
    while (!que.empty()) que.pop();
    que.push(start);
    memset(cnt, 0, sizeof(cnt));
    cnt[start] = 1;
    while (!que.empty())
    {
        int u = que.front(); que.pop();
        vis[u] = false;
        for (int i = 0; i<E[u].size(); i++)
        {
            int v = E[u][i].v;
            if (dist[v]>dist[u] + E[u][i].cost)
            {
                dist[v] = dist[u] + E[u][i].cost;
                if (!vis[v])
                {
                    vis[v] = true;
                    que.push(v);
                    if (++cnt[v] > n) return;
                }
            }
        }
    }
}

int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++)
            E[i].clear();
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &Map[i].a, &Map[i].b, &Map[i].c);
            addedge(Map[i].a, Map[i].b, Map[i].c);
        }
        int ans = 0;
        SPFA(1, n);
        for (int i = 1; i <= n; i++)
            ans += dist[i];

        for (int i = 1; i <= n; i++)
            E[i].clear();
        for (int i = 1; i <= m; i++)
            addedge(Map[i].b, Map[i].a, Map[i].c);
        SPFA(1, n);
        for (int i = 1; i <= n; i++)
            ans += dist[i];
        printf("%d\n", ans);
    }
    return 0;
}