• POJ_2387_最短路


    Til the Cows Come Home
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 46859   Accepted: 15941

    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


    熟悉一下刚学的SPFA,注意双向边,所以边数组要开倍。
    #include<iostream>
    #include<cstdio>
    #include<map>
    #include<cstring>
    #include<queue>
    using namespace std;
    #define V 1005
    #define E 4005
    #define INF 99999999
    
    struct Edge
    {
        int v,w;
        int next;
    } edge[E];
    
    int head[V];
    
    int cnte=0;
    void addEdge(int a,int b,int c)
    {
        edge[cnte].v=b;
        edge[cnte].w=c;
        edge[cnte].next=head[a];
        head[a]=cnte++;
    }
    
    int t,n;
    int dist[V],vis[V];
    void Spfa()
    {
        for(int i=0; i<=n; i++)
        {
            dist[i]=INF;
            vis[i]=0;
        }
        queue<int>q;
        q.push(1);
        dist[1]=0;
        vis[1]=1;
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int i=head[u]; i!=-1; i=edge[i].next)
            {
                int v=edge[i].v;
                if(dist[v]>dist[u]+edge[i].w)
                {
                    dist[v]=dist[u]+edge[i].w;
                    if(!vis[v])
                    {
                        vis[v]=1;
                        q.push(v);
                    }
                }
            }
            vis[u]=0;
        }
    }
    int main()
    {
        //cout<<INF<<endl;
        scanf("%d%d",&t,&n);
        memset(head,-1,sizeof(head));
        for(int i=0; i<t; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            addEdge(a,b,c);
            addEdge(b,a,c);
        }
        Spfa();
        printf("%d
    ",dist[n]);
        return 0;
    }
     

    #include<iostream>#include<cstdio>#include<map>#include<cstring>#include<queue>using namespace std;#define V 1005#define E 4005#define INF 99999999
    struct Edge{    int v,w;    int next;} edge[E];
    int head[V];
    int cnte=0;void addEdge(int a,int b,int c){    edge[cnte].v=b;    edge[cnte].w=c;    edge[cnte].next=head[a];    head[a]=cnte++;}
    int t,n;int dist[V],vis[V];void Spfa(){    for(int i=0; i<=n; i++)    {        dist[i]=INF;        vis[i]=0;    }    queue<int>q;    q.push(1);    dist[1]=0;    vis[1]=1;    while(!q.empty())    {        int u=q.front();        q.pop();        for(int i=head[u]; i!=-1; i=edge[i].next)        {            int v=edge[i].v;            if(dist[v]>dist[u]+edge[i].w)            {                dist[v]=dist[u]+edge[i].w;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }        vis[u]=0;    }}int main(){    //cout<<INF<<endl;    scanf("%d%d",&t,&n);    memset(head,-1,sizeof(head));    for(int i=0; i<t; i++)    {        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        addEdge(a,b,c);        addEdge(b,a,c);    }    Spfa();    printf("%d ",dist[n]);    return 0;}

  • 相关阅读:
    Wiin10 深色模式 暗色 省电 经济 护眼dark mode energy saving ecol
    云主机终端现实中文
    systemctl enable与systemctl start的区别
    centos技巧
    用国内的vps登录亚马逊的ec2
    小米 redmi note 8 打开开发者选项
    javascript数组、对象和Null的typeof同为object,区分解决办法
    组件中是否可以判断slot是否有内容?
    xlsx-style 行高设置
    修改xlsx-style 源码 解决报错
  • 原文地址:https://www.cnblogs.com/jasonlixuetao/p/6115335.html
Copyright © 2020-2023  润新知