Silver Cow Party
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 12913 |
|
Accepted: 5778 |
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
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.
题目是让求所有点到目标地点的最短路中的最大值。
思路：用两个二维数组分别存边的信息，两次dijkstra求得最短路。也可以用spfa求，速度很慢。
#include"stdio.h"
#include"string.h"
#define N 1005
const 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; }