Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
5 5 1 2 20 2 3 30 3 4 20 4 5 20 1 5 100Sample Output
90Hint
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
這道題目是找最短路,單源最短路徑的典型題目,運用了迪傑斯特拉算法。
#include<stdio.h>
#define Max 0x3fffffff
int map[1005][1005];
int dis[1005];
void dijkstra(int n)
{
int visit[1001]={0};
int min,i,j,k;
visit[1]=1;
for(i=1;i<n;++i)
{
min=Max;
k=1;
for(j=1;j<=n;++j)
{
if(!visit[j]&&min>dis[j])
{
min=dis[j];
k=j;
}
}
visit[k]=1;
for(j=1;j<=n;++j)
{
if(!visit[j]&&dis[j]>dis[k]+map[k][j])
dis[j]=dis[k]+map[k][j];
}
}
printf("%d\n",dis[n]);
}
int main()
{
int t,n,i,j,from,to,cost;
while(scanf("%d%d",&t,&n)!=EOF)
{
for(i=1;i<=n;++i)
{
map[i][i]=0;
for(j=1;j<i;++j)
map[i][j]=map[j][i]=Max;
}
for(i=1;i<=t;++i)
{
scanf("%d%d%d",&from,&to,&cost);
if(cost<map[from][to]) //可能有多條路,只記錄最短的
map[from][to]=map[to][from]=cost;
}
for(i=1;i<=n;++i)
dis[i]=map[1][i];
dijkstra(n);
}
}