POJ 2387 最短路径模板

    Til the Cows Come Home    Time Limit: 1000MS    Memory Limit: 65536K    Total Submissions: 60567    Accepted: 20615    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.    Source    USACO 2004 November    标准的最短路径题，dijkstra 算法的模板    踩到了两个坑，第一个，两点之间可能有不止一条路，所以要去除重复    第二个，我一般define MAXINT 喜欢用0x7fffffff  然而，只要这个值再加任意值，就炸了，所以最好还是用10000050 这类的吧#include<cstdio>#include<cstring>#define MAX 1000000#define MAXN 1005int map[MAXN][MAXN];int dis[MAXN];int visited[MAXN];void dijkstra(int n){    int k,min;    for(int i=1;i<=n;i++)    {        dis[i]=map[1][i];        visited[i]=0;     }    for(int i=1;i<=n;i++)    {        k=0;        min=MAX;        for(int j=1;j<=n;j++)        {            if(visited[j]==0&&dis[j]<min)            {                min=dis[j];                k=j;            }        }        visited[k]=1;        for(int j=1;j<=n;j++)        {            if(visited[j]==0&&dis[k]+map[k][j]<dis[j])                {                    dis[j]=dis[k]+map[k][j];                }        }    } }int main(){    int t,n,a,b,len;    scanf("%d%d",&t,&n);    memset(visited,0,sizeof(visited));    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)            map[i][j]=MAX;     for(int i=1;i<=t;i++)    {        scanf("%d%d%d",&a,&b,&len);        if(len<map[a][b])        {            map[a][b]=len;            map[b][a]=map[a][b];        }    }    dijkstra(n);    dis[1]=0;    printf("%d\n",dis[n]);  }