zoukankan      html  css  js  c++  java
  • 第k最短路A*启发式搜索

    Remmarguts' Date
    Time Limit: 4000MS   Memory Limit: 65536K
    Total Submissions: 21549   Accepted: 5862

    Description

    "Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks' head, he told them a story. 

    "Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission." 

    "Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)" 

    Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister's help! 

    DETAILS: UDF's capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince' current place. M muddy directed sideways connect some of the stations. Remmarguts' path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate. 

    Input

    The first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T. 

    The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).

    Output

    A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output "-1" (without quotes) instead.

    Sample Input

    2 2
    1 2 5
    2 1 4
    1 2 2
    

    Sample Output

    14
    #include"stdio.h"
    #include"string.h"
    #include"iostream"
    #include"map"
    #include"string"
    #include"queue"
    #include"stdlib.h"
    #include"algorithm"
    #include"math.h"
    #define M 1009
    #define eps 1e-10
    #define inf 100000000
    #define mod 100000000
    #define INF 0x3f3f3f3f
    using namespace std;
    struct Lnode
    {
        int u,v,w,next;
    }edge[M*300];
    int t,head[M],h[M],num[M],use[M];
    void init()
    {
        t=0;
        memset(head,-1,sizeof(head));
    }
    void add(int u,int v,int w)
    {
        edge[t].u=u;
        edge[t].v=v;
        edge[t].w=w;
        edge[t].next=head[u];
        head[u]=t++;
    }
    void dijstra(int n,int s)
    {
        int i,j;
        memset(use,0,sizeof(use));
        memset(h,INF,sizeof(h));
        h[s]=0;
        for(i=1;i<=n;i++)
        {
            int mini=INF;
            int tep=-1;
            for(j=1;j<=n;j++)
            {
                if(!use[j]&&h[j]<mini)
                {
                    mini=h[j];
                    tep=j;
                }
            }
            if(tep==-1)break;
            use[tep]=1;
            for(j=head[tep];j!=-1;j=edge[j].next)
            {
                int v=edge[j].v;
                if(h[v]>h[tep]+edge[j].w)
                    h[v]=h[tep]+edge[j].w;
            }
        }
    }
    struct node
    {
        int v,g,h;
        friend bool operator<(node a,node b)
        {
            return a.g+a.h>b.g+b.h;
        }
    };
    int bfs(int n,int s,int t,int k)
    {
        int i;
        priority_queue<node>q;
        memset(num,0,sizeof(num));
        node cur;
        cur.v=s;
        cur.g=0;
        cur.h=h[s];
        q.push(cur);
        while(!q.empty())
        {
            cur=q.top();
            q.pop();
            num[cur.v]++;
            if(num[cur.v]>k)continue;
            if(num[t]==k&&t==cur.v)return cur.g;
            for(i=head[cur.v];i!=-1;i=edge[i].next)
            {
                int v=edge[i].v;
                node now;
                now.v=v;
                now.g=cur.g+edge[i].w;
                now.h=h[v];
                q.push(now);
            }
        }
        return -1;
    }
    struct line
    {
        int a,b,c;
    }e[M*300];
    int main()
    {
        int n,m,i,start,endl,k;
        while(scanf("%d%d",&n,&m)!=-1)
        {
            init();
            for(i=1;i<=m;i++)
            {
                scanf("%d%d%d",&e[i].a,&e[i].b,&e[i].c);
                add(e[i].b,e[i].a,e[i].c);
            }
            scanf("%d%d%d",&start,&endl,&k);
            if(start==endl)k++;//注意的地方
            dijstra(n,endl);
            init();
            for(i=1;i<=m;i++)
                add(e[i].a,e[i].b,e[i].c);
            int ans=bfs(n,start,endl,k);
            printf("%d
    ",ans);
        }
        return 0;
    }
    


  • 相关阅读:
    TCP建立连接和断开连接
    城市三级联动Springmvc+mysql
    JS页面间数据传递的各种方法
    点击Button弹出登陆注册框
    一个多表联合查询引发的思考
    Linq扩展方法获取单个元素
    Visual Studio 常见问题
    格式化asp.net mvc视图页面
    2014上半年度挑战回顾
    2014年上半度年度挑战
  • 原文地址:https://www.cnblogs.com/mypsq/p/4348125.html
Copyright © 2011-2022 走看看