POJ3268 Silver Cow Party 【Dijkstra】

Silver Cow Party

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

因为是求一个来回，所以构造顺逆两张图，两次Dijkstra即可。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;

#define maxn 1010
#define maxm 100010
#define inf 0x3f3f3f3f

int N, M, X;
int head1[maxn], head2[maxn], id1, id2;
struct Node {
    int v, w, next;
} E1[maxm], E2[maxm];
int dist1[maxn], dist2[maxn];
bool vis1[maxn], vis2[maxn];

void addEdge(int head[], Node E[], int& id, int u, int v, int w) {
    E[id].v = v; E[id].w = w;
    E[id].next = head[u]; head[u] = id++;
}

void getMap() {
    int u, v, w; id1 = id2 = 1;
    while(M--) {
        scanf("%d%d%d", &u, &v, &w);
        addEdge(head1, E1, id1, u, v, w);
        addEdge(head2, E2, id2, v, u, w); // reverse
    }
}

int getNext(int dist[], bool vis[]) {
    int u = 0, i, v = inf;
    for(i = 1; i <= N; ++i)
        if(!vis[i] && dist[i] < v) {
            v = dist[i]; u = i;
        }
    return u;
}

void Dijkstra(int source, int head[], Node E[], int dist[], bool vis[]) {
    int i, u = source, v, w;
    memset(dist, 0x3f, sizeof(int) * (N + 1));
    dist[u] = 0;
    while(u) {
        vis[u] = 1;
        for(i = head[u]; i; i = E[i].next)
            dist[v=E[i].v] = min(dist[v], dist[u] + E[i].w);
        u = getNext(dist, vis);
    }
}

int main() {
    // freopen("stdin.txt", "r", stdin);
    scanf("%d%d%d", &N, &M, &X);
    getMap();
    Dijkstra(X, head1, E1, dist1, vis1);
    Dijkstra(X, head2, E2, dist2, vis2);
    int ans = 0;
    for(int i = 1; i <= N; ++i)
        ans = max(ans, dist1[i] + dist2[i]);
    printf("%d\n", ans);
    return 0;
}