学步园

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?

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.

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

10

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.

USACO 2007 February Silver
先在原图上求一次单源最短路，再把所有边反向，再求一次，把距离累加即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int maxn = 100000 + 5;
const int INF = 1000000000;
typedef long long LL;
typedef pair<int,int> P;

int n,m,x;
struct Edge{
    int from,to;
    LL dis;
}e[maxn];

LL ans[maxn];
LL d[maxn];
int vis[maxn];
vector<Edge> G[maxn];

void Dij(int x){
    priority_queue<P,vector<P>,greater<P> > Q;
    memset(vis,0,sizeof(vis));
    for(int i = 0;i <= n;i++) d[i] = INF;
    Q.push(P(0,x));
    while(!Q.empty()){
        P p = Q.top();Q.pop();
        int id = p.second;
        LL dis = p.first;
        if(vis[id] == 1) continue;
        vis[id] = 1;
        d[id] = dis;
        for(int i = 0;i < G[id].size();i++){
            Edge edgs = G[id][i];
            int to = edgs.to;
            LL der = edgs.dis;
            if(d[to] > d[id] + der){
                d[to] = d[id] + der;
                Q.push(P(d[to],to));
            }
        }
    }
}

int main(){
    while(scanf("%d%d%d",&n,&m,&x) != EOF){
        for(int i = 0;i <= n;i++) G[i].clear();
        for(int i = 0;i < m;i++){
            scanf("%d%d%I64d",&e[i].from,&e[i].to,&e[i].dis);
            G[e[i].from].push_back(e[i]);
        }
        Dij(x);
        for(int i = 0;i <= n;i++) ans[i] = d[i];
        for(int i = 0;i <= n;i++) G[i].clear();
        for(int i = 0;i < m;i++){
            swap(e[i].to,e[i].from);
            G[e[i].from].push_back(e[i]);
        }
        Dij(x);
        LL Max = 0;
        for(int i = 1;i <= n;i++){
            ans[i] += d[i];
            if(i != x){
                Max = max(Max,ans[i]);
            }
        }
        printf("%d\n",Max);
    }
    return 0;
}