Silver Cow Party

 

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14838		Accepted: 6711

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

正反最短路；

#include <iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
#define MAXN 1010
#define MAXM 100010
#define INF 0x3f3f3f3f
using namespace std;
struct node_z
{
int u,v,w;
int next;
}G_z[MAXM];
struct node_f
{
int u,v,w;
int next;
}G_f[MAXM];
int head_z[MAXN],head_f[MAXN];
int dis_f[MAXN],dis_z[MAXN];
bool vis[MAXN];
int n,m,x;
int k;
void build_z(int u,int v,int w)
{
G_z[k].u=u;
G_z[k].v=v;
G_z[k].w=w;
G_z[k].next=head_z[u];
head_z[u]=k;

}
void build_f(int u,int v,int w)
{
G_f[k].u=u;
G_f[k].v=v;
G_f[k].w=w;
G_f[k].next=head_f[u];
head_f[u]=k;
}
void build()
{
k=0;
int u,v,w;
memset(head_f,-1,sizeof(head_f));
memset(head_z,-1,sizeof(head_z));
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
build_z(u,v,w);
build_f(v,u,w);
k++;
}
}
void SPFA_f(int s)
{
for(int i=1;i<=n;i++)
{
dis_f[i]=INF;
vis[i]=0;
}
dis_f[s]=0;
queue<int>Q;
Q.push(s);
vis[s]=1;
int u,v,g;
while(!Q.empty())
{
u=Q.front();Q.pop();vis[u]=0;
for(int i=head_f[u];i!=-1;i=G_f[i].next)
{
g=dis_f[u]+G_f[i].w;v=G_f[i].v;
if(g<dis_f[v])
{
dis_f[v]=g;
if(!vis[v])
{
Q.push(v);
vis[v]=1;
}
}
}
}
}
void SPFA_z(int s)
{
for(int i=1;i<=n;i++)
{
dis_z[i]=INF;
vis[i]=0;
}
dis_z[s]=0;
queue<int>Q;
Q.push(s);
vis[s]=1;
int u,v,g;
while(!Q.empty())
{
u=Q.front();Q.pop();vis[u]=0;
for(int i=head_z[u];i!=-1;i=G_z[i].next)
{
g=dis_z[u]+G_z[i].w;v=G_z[i].v;
if(g<dis_z[v])
{
dis_z[v]=g;
if(!vis[v])
{
Q.push(v);
vis[v]=1;
}
}
}
}
}
int main()
{

while(~scanf("%d%d%d",&n,&m,&x))
{
build();
SPFA_z(x);
SPFA_f(x);
int ans=dis_f[1]+dis_z[1],g;
for(int i=2;i<=n;i++)
if((g=dis_f[i]+dis_z[i])>ans)
ans=g;
printf("%d\n",ans);
}
return 0;
}