poj3268(Silver Cow Party)最短路

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.

题意：求所有节点到源点再返回各自节点的最短路的最大值

题解：最短路问题，构建两张图，第二张是所有边方向取反，各做一次dijkstra就行

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxv = 1010;

struct edge
{
    int v, d;
    bool operator < (const edge& b) const
    {
        return d > b.d;
    }
};

vector<edge> G1[maxv], G2[maxv];
int d1[maxv], d2[maxv], vis[maxv];
int n, m, s;

void dijkstra1(int s)
{
    memset(d1, INF, sizeof(d1));
    memset(vis, 0, sizeof(vis));
    d1[s] = 0;
    priority_queue<edge> q;
    q.push({s, 0});
    while (!q.empty())
    {
        edge e = q.top();
        q.pop();
        int from = e.v;
        if (vis[from])
            continue;
        vis[from] = 1;
        for (int i = 0; i < G1[from].size(); ++i)
        {
            int to = G1[from][i].v, d = G1[from][i].d;
            if (d1[to] > d1[from] + d)
            {
                d1[to] = d1[from] + d;
                q.push({to, d1[to]});
            }
        }
    }
}

void dijkstra2(int s)
{
    memset(d2, INF, sizeof(d2));
    memset(vis, 0, sizeof(vis));
    d2[s] = 0;
    priority_queue<edge> q;
    q.push({s, 0});
    while (!q.empty())
    {
        edge e = q.top();
        q.pop();
        int from = e.v;
        if (vis[from])
            continue;
        vis[from] = 1;
        for (int i = 0; i < G2[from].size(); ++i)
        {
            int to = G2[from][i].v, d = G2[from][i].d;
            if (d2[to] > d2[from] + d)
            {
                d2[to] = d2[from] + d;
                q.push({to, d2[to]});
            }
        }
    }
}

int main()
{
    cin >> n >> m >> s;
    while (m--)
    {
        int a, b, d;
        scanf("%d%d%d", &a, &b, &d);
        G1[a-1].push_back({b-1, d});
        G2[b-1].push_back({a-1, d});
    }
    dijkstra1(s-1);
    dijkstra2(s-1);
    int ans = -1;
    for (int i = 0; i < n; ++i)
        ans = max(ans, d1[i]+d2[i]);
    cout << ans << endl;
    return 0;
}