POJ 3268 Silver Cow Party

http://poj.org/problem?id=3268

有一个点X开party 求其他所有点  去的路程和返程的和的最小值中的最大值

这里有一个小技巧 

总是把X当做原点 

返程是正常的最短路

去程就把路径翻转 求最短路 

这样只需要两次求最短路 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#define MAXV 1007
#define MAXE 100007
#define INF 0x3f3f3f3f
using namespace std;

typedef pair<int,int> P ;//first 是距离 second是编号
struct Edge
{
    int to, cost, next;
    Edge() {}
    Edge(int to, int cost, int next) : to(to), cost(cost), next(next) {}
}edge1[MAXE], edge2[MAXE];

int head1[MAXV], head2[MAXV];
int num1 = 0, num2 = 0;
void Add1(int from, int to, int cost)
{
    edge1[num1] = Edge(to, cost, head1[from]);
    head1[from] = num1++;
}

void Add2(int from, int to, int cost)
{
    edge2[num2] = Edge(to, cost, head2[from]);
    head2[from] = num2++;
}

int N, M, X;

int dijkstra(int s,int dist[], int head[], Edge edge[])
{
    //int dist[MAXV];
    priority_queue<P> que;
    fill(dist, dist+MAXV, INF);
    dist[s] = 0;
    que.push(P(0, s));
    while (!que.empty())
    {
        P p = que.top();
        que.pop();
        int t = head[p.second];
        if (dist[p.second] < p.first) continue;
        while (t != -1)
        {
            Edge e = edge[t];
            if(dist[e.to] > dist[p.second] + e.cost)
            {
                dist[e.to] = dist[p.second] + e.cost;
                que.push(P(dist[e.to], e.to));
            }
            t = e.next;
        }
    }
}

int main()
{
    int ans[MAXV];
    int dist1[MAXV], dist2[MAXV];
    freopen("in.txt", "r", stdin);
    scanf("%d%d%d", &N, &M, &X);
    memset(head1, -1, sizeof(head1));
    memset(head2, -1, sizeof(head2));
    memset(edge1, -1, sizeof(edge1));
    memset(edge2, -1, sizeof(edge2));
    memset(ans, 0, sizeof(ans));
    for (int i = 0; i < M; i++)
    {
        int from, to, cost;
        scanf("%d%d%d", &from, &to, &cost);
        Add1(from, to, cost);
        Add2(to, from, cost);
    }
    dijkstra(X, dist1, head1, edge1);//正常返回的路径
    dijkstra(X, dist2, head2, edge2);//路径翻转之后 得到的dist就是 i到达X的路径
    for (int i = 1; i <= N; i++)
    {
        ans[i] = dist1[i] + dist2[i];
    }
    sort(ans+1, ans+N+1);
    printf("%d\n", ans[N]);
    return 0;
}