POJ 3268 Silver Cow Party 最短路

原题链接：http://poj.org/problem?id=3268

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15545		Accepted: 7053

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.

Sample Input

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

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

Source

USACO 2007 February Silver

题意

有一只牛举办派对，其他的牛去参加，牛都会走最短路，并且派对结束还要回到自己家里。问哪头牛走的路径最长，输出最长路径。

题解

就跑两边spfa，正着反着跑两次。然后就搞定了。

代码

#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cstdio>
#define INF 1000006
#define MAX_N 1003
using namespace std;

struct node {
public:
    int u, c;

    node(int uu, int cc) : u(uu), c(cc) { }

    node() { }
};

struct edge {
public:
    int to, cost;

    edge(int t, int c) : to(t), cost(c) { }

    edge() { }
};

queue<node> que;
int n,m,x;
void spfa(int s,vector<edge> G[],int d[]) {
    fill(d, d + n + 1, INF);
    que.push(node(s, 0));
    d[s] = 0;
    while (que.size()) {
        node now = que.front();
        que.pop();
        if (now.c != d[now.u])continue;
        int u = now.u;
        for (int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].to;
            int t = d[u] + G[u][i].cost;
            if (t < d[v]) {
                d[v] = t;
                que.push(node(v, t));
            }
        }
    }
}

vector<edge> G[MAX_N],rG[MAX_N];
int d[MAX_N],rd[MAX_N];
int main() {
    scanf("%d%d%d", &n, &m, &x);
    for (int i = 0; i < m; i++) {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        G[u].push_back(edge(v, c));
        rG[v].push_back(edge(u, c));
    }
    spfa(x, G, d);
    while (que.size())que.pop();
    spfa(x, rG, rd);
    int ans = 0;
    for (int i = 1; i <= n; i++)ans = max(ans, d[i] + rd[i]);
    cout<<ans<<endl;
    return 0;
}