<poj

　　本题链接 ： http://poj.org/problem?id=3268

　　题目大意：牛们要去聚会，输入N = 顶点数（牛场）；M = 边（路）的数目； X = 终点 （聚会点）。问题：求来回时间的最大值。

Description:

　One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input：

Line 1: Three space-separated integers, respectively: N, M, and X 
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output：

Line 1: One integer: the maximum of time any one cow must walk.

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output：

10
　
　解题思路：因为本题是单向的，所以在做的时候可以做一个正向图和一个反向图，分别求解来回的时间，然后找和的最大值。

这里是我的代码：

#include <cstring>
#include <iostream>
#define INF 9999999
using namespace std;

bool used[100005];
int V, E;
const int maxn = 1005;

void dijkstra (int s, int cost[][1005], int d[]) {
    fill (d, d + V + 1, INF);
    fill (used,used + V + 1,false);
    d[s] = 0;
    while (true) {
        int v = -1;
        for (int u = 1; u <= V; u++) {
            if (!used[u] && (v == -1 || d[u] < d[v])) v = u;
        }
        if (v == -1) break;
        used[v] = true;
        for (int u = 1; u <= V; ++u) {
            if (d[u] >  d[v] + cost[v][u]) {
                d[u] = d[v] + cost[v][u];
            }
        }
    }
}

int cost[maxn][maxn];
int rcost[maxn][maxn];

int main () {
    int d[maxn];
    int rd[maxn];
    int x, y, w;
    int sum[maxn];
    int S;
    cin >> V >> E >> S;
    
    for (int i = 1;i <= V; ++i)
        for (int j = 1; j <= V; ++j)
            rcost[i][j] = cost[i][j] = INF;
    
    for (int i = 0; i < E; ++i) {
        cin >> x >> y >> w;
        rcost[y][x] = cost[x][y] = w;
    }
    
    dijkstra(S, cost, d);//分别计算最短路径
    dijkstra(S, rcost, rd);
    
    for (int i = 1; i <= V; ++i)//求和
        sum[i] = d[i] + rd[i];
    
    int maxnum = sum[1];
    for (int i = 1; i <= V; ++i)///找最大值
        if (sum[i] > maxnum)
            maxnum = sum[i];
    
    cout << maxnum << endl;
    
    return 0;
}

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