Silver Cow Party - POJ 3268 spfa

最新推荐文章于 2021-05-04 13:37:00 发布

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本文详细介绍了使用SPFA算法求解有向图中每只牛从农场到特定农场参加聚会并返回的最短路径，并通过实例展示了算法的应用。重点在于理解SPFA算法的工作原理和其在解决实际问题中的优势。

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13006		Accepted: 5824

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.

题意：在一个有向图中，问每只牛到X再回来的最短距离的最大值。

思路：练习下spfa吧。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{
    int y,w,next;
}e[100010];
int n,m,h[1010],d[1010],x[100010],y[100010],w[100010],tot,ans,f[1010];
bool vis[1010];
queue<int> qu;
void ins(int x,int y,int w)
{
    e[++tot].y=y;
    e[tot].w=w;
    e[tot].next=h[x];
    h[x]=tot;
}
void spfa(int s)
{
    int i,j,k,x,y;
    qu.push(s);
    memset(vis,0,sizeof(vis));
    for(i=1;i<=n;i++)
       d[i]=1000000000;
    d[s]=0;
    vis[s]=true;
    while(!qu.empty())
    {
        x=qu.front();
        qu.pop();
        for(i=h[x];i;i=e[i].next)
        {
            y=e[i].y;
            if(d[x]+e[i].w<d[y])
            {
                d[y]=d[x]+e[i].w;
                if(!vis[y])
                {
                    vis[y]=1;
                    qu.push(y);
                }
            }
        }
        vis[x]=0;
    }
}
int main()
{
    int i,j,k;
    scanf("%d%d%d",&n,&m,&k);
    ans=0;
    tot=0;
    memset(h,0,sizeof(h));
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x[i],&y[i],&w[i]);
        ins(x[i],y[i],w[i]);
    }
    spfa(k);
    for(i=1;i<=n;i++)
       f[i]+=d[i];
    tot=0;
    memset(h,0,sizeof(h));
    for(i=1;i<=m;i++)
       ins(y[i],x[i],w[i]);
    spfa(k);
    for(i=1;i<=n;i++)
    {
        f[i]+=d[i];
        ans=max(ans,f[i]);
    }
    printf("%d\n",ans);
}