POJ 3268 Silver Cow Party （最短路径）
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
Http
POJ:https://vjudge.net/problem/POJ-3268
Source
最短路径
题目大意
在一个有向图中，求所有点都走到一个点再走回来的最短距离中的最大值
解决思路
我们知道单源最短路的求法，即从一个点走到其他点，那么我们只要把有向图中的边反过来求一遍就是从其他点走到一个点的最短距离
这里我们用spfa解决
代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;const int maxN=1001;
const int inf=2147483647;class Edge
{
public:int v,w;
};int n,m,X;
vector<Edge> E1[maxN];
vector<Edge> E2[maxN];
queue<int> Q;
bool inqueue[maxN];
int Dist1[maxN];
int Dist2[maxN];int main()
{scanf("%d%d%d",&n,&m,&X);for (int i=1;i<=m;i++){int u,v,w;scanf("%d%d%D",&u,&v,&w);E1[u].push_back((Edge){v,w});//存正图E2[v].push_back((Edge){u,w});//存反图}memset(Dist1,127,sizeof(Dist1));//第一遍spfamemset(inqueue,0,sizeof(inqueue));Dist1[X]=0;inqueue[X]=1;while (!Q.empty())Q.pop();Q.push(X);do{int u=Q.front();Q.pop();inqueue[u]=0;for (int i=0;i<E1[u].size();i++){int v=E1[u][i].v;int w=E1[u][i].w;if (Dist1[v]>Dist1[u]+w){Dist1[v]=Dist1[u]+w;if (inqueue[v]==0){Q.push(v);inqueue[v]=1;}}}}while (!Q.empty());memset(Dist2,127,sizeof(Dist2));//第二遍spfamemset(inqueue,0,sizeof(inqueue));Dist2[X]=0;inqueue[X]=1;Q.push(X);do{int u=Q.front();Q.pop();inqueue[u]=0;for (int i=0;i<E2[u].size();i++){int v=E2[u][i].v;int w=E2[u][i].w;if (Dist2[v]>Dist2[u]+w){Dist2[v]=Dist2[u]+w;if (inqueue[v]==0){Q.push(v);inqueue[v]=1;}}}}while (!Q.empty());int Ans=0;for (int i=1;i<=n;i++)Ans=max(Ans,Dist1[i]+Dist2[i]);//统计最大值cout<<Ans<<endl;return 0;
}
)
