# 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.

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

10

## 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spfa
memset(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spfa
memset(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;
}

posted @ 2017-07-23 16:45  SYCstudio  阅读(...)  评论(...编辑  收藏