poj 3268 （最短路 spfa）

最新推荐文章于 2019-05-13 21:22:49 发布

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本文介绍了一种用于寻找图中两点间最短路径的SPFA算法，并通过一个具体问题实例展示了如何利用该算法解决复杂的路径寻优问题。文章首先概述了问题背景——即如何计算各农场到聚会地点并返回的最大所需时间，接着详细解释了SPFA算法的基本原理及其应用过程。

Language:

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12910		Accepted: 5776

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.

Source

USACO 2007 February Silver

题意：4个村庄，8条有向边，2为起点，问其他村庄到2后又回去，求哪一个村庄浪费的时间最多，只要输出时间

spfa算法其实挺还好，关键是里面的邻接表不好理解

思路：两次spfa，第一次算出别的村庄来的时间，第二次反向建边，反向建边是说到达此点的点有哪些然后更新那些点，

在一次强调，里面的邻接表不好理解，献血邻接表再学spfa

直接上代码了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define N 10005
#define maxe 0x3f3f3f3f

int n,m,x;
int vis[N],dis[N];
int head[N];
int num;
int temp[N];

struct stud1{
int s,e,w;
}ss[N*2];

struct stud{
int to,w,next;
}ee[N*2];

void build(int u,int v,int w)
{
    ee[num].to=v;
    ee[num].w=w;
    ee[num].next=head[u];
    head[u]=num;
    num++;
}

void spfa(int x)
{
    int i,j;
    for(i=1;i<=n;i++)
    {
        dis[i]=maxe;
        vis[i]=0;
    }

    dis[x]=0;
    vis[x]=1;

    queue<int>q;
    q.push(x);

    int cur;
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        vis[cur]=0;

        for(i=head[cur];i!=-1;i=ee[i].next)
        {
            if(dis[ee[i].to]>dis[cur]+ee[i].w)
            {
                dis[ee[i].to]=dis[cur]+ee[i].w;
                if(!vis[ee[i].to])
                {
                    q.push(ee[i].to);
                    vis[ee[i].to]=1;
                }
            }
        }
    }

    for(i=1;i<=n;i++)
       temp[i]+=dis[i];
}

int main()
{
    int i,j;
    while(~scanf("%d%d%d",&n,&m,&x))
    {
     num=0;
     for(i=1;i<=n;i++)
     {
         temp[i]=0;
         head[i]=-1;
     }

     for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&ss[i].s,&ss[i].e,&ss[i].w);
            build(ss[i].s,ss[i].e,ss[i].w);
        }

        spfa(x);

        num=0;
       for(i=1;i<=n;i++)
        head[i]=-1;

        for(i=0;i<m;i++)
          build(ss[i].e,ss[i].s,ss[i].w);

        spfa(x);

        j=1;
        for(i=2;i<=n;i++)
            if(temp[i]>temp[j])
            j=i;
        printf("%d\n",temp[j]);
    }
    return 0;
}