POJ-3268-Silver Cow Party（SPFA 反向建图）

最新推荐文章于 2022-03-13 20:30:09 发布

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最新推荐文章于 2022-03-13 20:30:09 发布

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分类专栏：最短路径文章标签： SPFA

本文链接：https://blog.youkuaiyun.com/qq_32680617/article/details/52179420

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本文介绍了一道经典的最短路径问题，通过两遍SPFA算法解决N个农场到特定聚会地点并返回的最短时间问题。利用正反向建图优化算法效率。

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Silver Cow Party
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 18662 Accepted: 8536
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 Ti (1 ≤ Ti ≤ 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: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti 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：N个点，M条边，起点为X，求出每个点在有向图中来回X的最小距离中的最大距离

代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
//两遍SPFA
//反向建图
const int maxn=100005;
const int INF=0x3f3f3f3f;
int u[maxn];
int v[maxn];
int w[maxn];
int first[maxn];
int next[maxn];
int dis[maxn];
int vis[maxn];//标记是否在队列中
int N,M,X;
void init_map()//正向建图
{
    for(int i=1; i<=N; i++)
    {
        dis[i]=INF;
        vis[i]=0;
        first[i]=-1;
    }
    for(int i=1; i<=M; i++)
    {
        scanf("%d%d%d",&u[i],&v[i],&w[i]);
        next[i]=first[u[i]];
        first[u[i]]=i;
    }
}
void set_map()//反向建图
{
    for(int i=1; i<=N; i++)
    {
        dis[i]=INF;
        vis[i]=0;
        first[i]=-1;
    }
    for(int i=1; i<=M; i++)
    {
        next[i]=first[v[i]];
        first[v[i]]=i;
    }
}
void SPFA(int star,bool flag)//flag标记是正向还是反向
{
    queue<int>q;
    q.push(star);
    dis[star]=0;
    vis[star]=1;
    while(!q.empty())
    {
        star=q.front();
        q.pop();
        vis[star]=0;
        for(int k=first[star]; k!=-1; k=next[k])
        {
            int temp=(flag==1?v[k]:u[k]);
            if(dis[temp]>dis[star]+w[k])
            {
                dis[temp]=dis[star]+w[k];
                if(vis[temp]==0)
                {
                    q.push(temp);
                    vis[temp]=1;
                }
            }
        }
    }
}
int main()
{
    scanf("%d%d%d",&N,&M,&X);
    init_map();
    SPFA(X,1);
    int max_num=0;
    int flag[1005];
    for(int i=1; i<=N; i++)
        flag[i]=dis[i];
    set_map();
    SPFA(X,0);
    for(int i=1; i<=N; i++)
        max_num=max(max_num,dis[i]+flag[i]);
    printf("%d\n",max_num);
    return 0;
}