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最新推荐文章于 2021-04-01 18:48:34 发布

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本文探讨了在一个由N个农场组成的网络中，每头牛从各自农场前往并返回大型聚会地点的最短时间路径问题。利用Dijkstra算法变形处理单向图，寻找所有牛往返所需时间的最大值。通过预先计算从聚会地点到各农场的最短路径，避免了重复计算，提高了效率。

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.

Dijkstra算法变形求最短路径中的最大值而且是单向图，并且最短路径是一去一回，这个题搞得我头都要炸一直超时，应该先打好表在直接找就好了。每次都是到X这个点(有进有出)不如就全部看成从X到所有的点的最短路径，算回去的路的时候把单向图转置一下就好了
来和回的最短路径分别存到两个数组然后直接暴力。

#include <algorithm>
#include <iostream>
#include<string.h>
#include <stdio.h>
#include<cmath>
using namespace std;
int dp[19009],a[4000][1009],b[4000],n,a1[4000],a2[4000],x;

//vector<int>b;
void roda(int *b)
{
        memset(dp,0,sizeof(dp));
        for(int j=1; j<=n; j++)
            b[j]=a[x][j];
        b[x]=0;
        dp[x]=1;
        for(int j=1; j<n; j++)
        {
            int maxx=0x3f3f3f3f,u=-1;
            for(int k=1; k<=n; k++)
            {
                if(maxx>b[k]&&!dp[k])
                {
                    maxx=b[k];
                    u=k;
                }
            }
            if(u==-1)
                break;
            dp[u]=1;
            for(int o=1; o<=n; o++)
            {
                if(b[o]>b[u]+a[u][o]&&o!=x)
                    b[o]=b[u]+a[u][o];
            }
        }
}
void tran()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<i;j++)
        {
            swap(a[i][j],a[j][i]);
        }
}
int main()
{
    int m,s,flag=0,g,h,j;
    while(~scanf("%d%d%d",&n,&m,&x))
    {
        memset(a,0x3f3f3f3f,sizeof(a));
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&g,&h,&j);
            a[g][h]=j;
        }
        roda(a1);
        tran();
        roda(a2);
        int maxx=-1;
        for(int i=1; i<=n; i++)
                maxx=max(a1[i]+a2[i],maxx);
        cout<<maxx<<endl;

    }

}