题目链接：http://poj.org/problem?id=2387
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 45414 | Accepted: 15405 |
Description
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.
Input
* 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.
Output
Sample Input
5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100
Sample Output
90
题目大意：有N个节点，T条路（带权值），求1到N的最短路径。
解题思路：最短路迪杰斯特拉算法（模板题）。
AC代码：
#include <stdio.h>
#include <string.h>
#define inf 999999999
int visit[];
int dis[];
int p[][];
int n;
int dijkstra()
{
int i,j,pos = ,minn;
memset(visit,,sizeof(visit));
visit[] = ;
dis[] = ;
for (i = ; i <= n; i ++)
dis[i] = p[][i];
for (i = ; i <= n; i ++)
{
minn = inf;
for (j = ; j <= n; j ++)
{
if (!visit[j] && dis[j] < minn)
{
minn = dis[j];
pos = j;
}
}
visit[pos] = ;
for (j = ; j <= n; j ++)
if (!visit[j] && dis[j] > dis[pos]+p[pos][j])
dis[j] = dis[pos]+p[pos][j];
}
return dis[n];
}
int main ()
{
int t,a,b,c,i,j;
while (~scanf("%d%d",&t,&n))
{
for (i = ; i <= n; i ++)
for (j = ; j <= n; j ++)
p[i][j] = inf; for (i = ; i < t; i ++)
{
scanf("%d%d%d",&a,&b,&c);
if (p[a][b] > c)
p[a][b] = p[b][a] = c;
}
printf("%d\n",dijkstra());
}
return ;
}