TOJ1693（Silver Cow Party）

Silver Cow Party  

Time Limit(Common/Java):2000MS/20000MS     Memory Limit:65536KByte
Total Submit: 58            Accepted: 28

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 farmB_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 February 2007

思路：对原图求一次最短路，再对反图进行一次最短路，然后找最求解即可.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
using namespace std;
const int maxn = 1100;
const int maxm = 101000;
const int INF = 0x7fffffff;
vector<pair<int, int> > g[maxn];
vector<pair<int, int> > rg[maxn];//reverse;
int gn, gm, x;
bool inque[maxn];
queue<int> Q;
int d[maxn];//保存原图最短径.
int rd[maxn];//保存反图的最短路径值.

void spfa2(int s) {
    int i;
    memset(inque, false, sizeof(inque));
    for(i = 1; i < maxn; i++) rd[i] = INF;
    rd[s] = 0;
    while(!Q.empty()) Q.pop();
    Q.push(s);
    inque[s] = true;
    while(!Q.empty()) {
        int u = Q.front();
        Q.pop();
        for(i = 0; i < (int)rg[u].size(); i++) {
            int t = rg[u][i].first;
            if( rd[u] != INF && rd[u] + rg[u][i].second < rd[t]) {
                rd[t] = rd[u] + rg[u][i].second;
                if(!inque[t]) {
                    inque[t] = true;
                    Q.push(t);
                }
            }
        }
        inque[u] = false;
    }
}

void work(const int s) {
    int i;
    int maxv = -1;
    for(i = 1; i <= gn; i++) {
        if(i != s) {
            if(d[i] + rd[i] > maxv) {
                maxv = d[i] + rd[i];
            }
        }
    }
    printf("%d\n", maxv);
}

void spfa1(int s) {
    int i;
    memset(inque, false, sizeof(inque));
    for(i = 1; i < maxn; i++) d[i] = INF;
    d[s] = 0;
    while(!Q.empty()) Q.pop();
    Q.push(s);
    inque[s] = true;
    while(!Q.empty()) {
        int u = Q.front();
        Q.pop();
        for(i = 0; i < (int)g[u].size(); i++) {
            int t = g[u][i].first;
            if(d[u] != INF && d[u] + g[u][i].second < d[t]) {
                d[t] = d[u] + g[u][i].second;
                if(!inque[t]) {
                    inque[t] = true;
                    Q.push(t);
                }
            }
        }
        inque[u] = false;
    }
}

int main()
{
    int i;
    int x, y, w;
    int start;
    pair<int, int> t;
    while(scanf("%d%d%d", &gn, &gm, &start) != EOF) {
        for(i = 1; i <= gn; i++) {
            g[i].clear();
            rg[i].clear();
        }
        for(i = 0; i < gm; i++) {
            scanf("%d%d%d", &x, &y, &w);
            t.first = y;
            t.second = w;
            g[x].push_back(t);
            t.first = x;
            t.second = w;
            rg[y].push_back(t);
        }
        spfa1(start);
        spfa2(start);
        work(start);
    }
    return 0;
}