Poj 3268 Silver Cow Party (Shortest Path)

Poj 3268 Silver Cow Party (Shortest Path)

Silver Cow Party

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:12913		Accepted:5778

Description

One cow from eachNFarms (1 ≤N≤ 1000) conveniently numbered 1 ..NIs going to attend the big cow party to be held at farm #X(1 ≤X≤N). A totalM(1 ≤MLess than or equal to 100,000) unidirectional (one-way roads connects pairs of farms; roadIRequiresTi(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 IWith three space-separated integers: Ai, Bi, And Ti. The described road runs from farm AiTo farm Bi, Requiring TiTime units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 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.

The question is to find the maximum value of the shortest path from all points to the target location.

Idea: Use two-dimensional arrays to store edge information separately, and use dijkstra twice to obtain the shortest path. You can also use spfa to calculate the speed, which is very slow.

#include"stdio.h"#include"string.h"#define N 1005const int inf=0x7fffffff;int n,m,x;int g1[N][N],g2[N][N];int dis1[N],dis2[N],mark[N];void dijkstra(int g[N][N],int *dis){    int i,u,min;    for(i=1;i<=n;i++)    {        dis[i]=inf;        mark[i]=0;    }    dis[x]=0;    while(1)    {        u=0;        min=inf;        for(i=1;i<=n;i++)        {            if(min>dis[i]&&!mark[i])            {                min=dis[i];                u=i;            }        }        if(u==0)            break;        mark[u]=1;        for(i=1;i<=n;i++)        {            if(g[u][i]
 
  dis[u]+g[u][i])                dis[i]=dis[u]+g[u][i];        }    }}int main(){    int i,j,u,v,w;    while(scanf("%d%d%d",&n,&m,&x)!=-1)    {        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)                g1[i][j]=g2[i][j]=inf;        }        for(i=0;i
  
   t?ans:t;        }        printf("%d\n",ans);    }    return 0;}