POJ 3268 Silver Cow Party

最新推荐文章于 2021-05-04 13:37:00 发布

原创最新推荐文章于 2021-05-04 13:37:00 发布 · 156 阅读

0 ·

CC 4.0 BY-SA版权

最短路问题专栏收录该内容

19 篇文章

订阅专栏

本文介绍了一种利用最短路径算法解决农场聚会中奶牛行走时间问题的方法。通过两次运行Dijkstra算法，分别计算各农场到聚会地点及返回的最短时间，并以此找出所有奶牛往返所需最长总时间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:25707		Accepted: 11729

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

Sample Output

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.

路是单向，从i到j在回来,即是从j到i,然后将道路反向，
再求一次j到i，两者相加即可。
代码：

#include<cstdio>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
int n,m,st,d[1003];
bool vis[1003];
using namespace std;
struct node
{
    int to,cost;
};
vector<node>go[1003],goback[1003];
struct node2
{
    int dec,len;
};
bool operator<(node2 a,node2 b)
{
    return a.len>b.len;
}
priority_queue<node2>P;
void Dija(vector<node> *rode)
{
    d[st]=0;
    node2 r;
    r.dec=st,r.len=0;
    P.push(r);
    while(!P.empty())
    {
        r=P.top();
        P.pop();
        int v=r.dec;
        if(d[v]<r.len)continue;
        vis[v]=1;
        int L=rode[v].size();
        for(int i=0; i<L; i++)
        {
            node e=rode[v][i];
            if(d[e.to]>d[v]+e.cost)
            {
                d[e.to]=d[v]+e.cost;
                node2 u;
                u.dec=e.to;
                u.len=d[e.to];
                P.push(u);
            }
        }
    }
}
int main()
{
    scanf("%d%d%d",&n,&m,&st);
    int i,j;
    memset(d,INF,sizeof(d));
    memset(vis,0,sizeof(vis));
    for(i=1; i<=m; i++)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        node r;
        r.to=y;
        r.cost=z;
        go[x].push_back(r);
        r.to=x;
        goback[y].push_back(r);
    }
    Dija(go);
    int d1[1003];
    for(i=1; i<=n; i++)
        d1[i]=d[i];
    memset(d,INF,sizeof(d));
    memset(vis,0,sizeof(vis));
    Dija(goback);
    int ma=0;
    for(i=1; i<=n; i++)
    {
        d1[i]+=d[i];
        ma=max(ma,d1[i]);
    }
    printf("%d\n",ma);
    return 0;
}