POJ 2387 Til the Cows Come Home【最短路】

题目来戳呀

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

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

• Line 1: Two integers: T and N

• 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

• Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90

Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.
题意：小羊想快快回家，输出最短距离+_+
想法：dijkstra模板题，就是想看看自己能不能完整的把dijkstra敲出来（还是暴露出一些小问题的′⌒`）
但是dijkstra算法很好理解啊，和最小生成树的prim算法几乎一样嘛^__^

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<string>using namespace std;const int inf=0x3f3f3f3f;const int maxn=1005;int vis[maxn];int temp,n,m;int dis[maxn],arr[maxn][maxn];void dijkstra(int v){    int minx;    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;++i)        dis[i]=arr[v][i];//下标错了    for(int i=1;i<=n;++i)    {        minx=inf;        temp=0;        for(int j=1;j<=n;++j)        {            if(!vis[j]&&dis[j]<minx)            {                minx=dis[j];                temp=j;            }        }        vis[temp]=1;        for(int j=1;j<=n;++j)        {            if(!vis[j]&&dis[temp]+arr[temp][j]<dis[j])                dis[j]=dis[temp]+arr[temp][j];        }    }}int main(){    while(~scanf("%d %d",&m,&n))    {        for(int i=1;i<=n;++i)        {            for(int j=1;j<=n;++j)            {                if(i==j)arr[i][j]=0;                else  arr[i][j]=inf;            }        }        while(m--)        {            int a,b;            scanf("%d %d %d",&a,&b,&temp);            if(temp<arr[a][b])                arr[a][b]=arr[b][a]=temp;        }        dijkstra(1);        printf("%d\n",dis[n]);    }    return 0;}

ps：寒假训练时的题，当时对最短路一无所知（捂脸）