zoukankan      html  css  js  c++  java
  • HDOJ1142 A Walk Through the Forest

    A Walk Through the Forest

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3140    Accepted Submission(s): 1147


    Problem Description
    Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.
    The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.
     
    Input
    Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.
     
    Output
    For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647
     
    Sample Input
    5 6 1 3 2 1 4 2 3 4 3 1 5 12 4 2 34 5 2 24 7 8 1 3 1 1 4 1 3 7 1 7 4 1 7 5 1 6 7 1 5 2 1 6 2 1 0
     
    Sample Output
    2 4
     1 //最短路径+记忆化搜索
     2 #include<cstdio>
     3 #include<cstring> 
     4 int map[1001][1001];
     5 int dis[1001];
     6 int dp[1001];
     7 
     8 void dijkstra(int v,int n)
     9 {
    10     bool visit[1001];
    11     memset(visit,false,sizeof(visit));
    12     int i,j,k,min;
    13     visit[v]=true;
    14     for(i=1;i<n;++i)
    15     {
    16         min=0x3fffffff;
    17         k=1;
    18         for(j=1;j<=n;++j)
    19         {
    20             if(!visit[j]&&dis[j]<min)
    21             {
    22                 min=dis[j];
    23                 k=j;
    24             }
    25         }
    26         visit[k]=true;
    27         for(j=1;j<=n;++j)
    28         {
    29             if(!visit[j]&&dis[j]>map[k][j]+min)
    30                 dis[j]=map[k][j]+min;
    31         }
    32     }
    33 }
    34 
    35 int mDFS(int v,int n)//记忆化搜索
    36 {
    37     if(dp[v]!=-1)   //dp[i]中记录的是i到终点的所有可能路径的数量
    38         return dp[v];
    39     if(v==2)
    40         return 1;
    41     dp[v]=0;
    42     for(int i=1;i<=n;++i)
    43         if(map[i][v]!=0x3fffffff&&dis[i]<dis[v]) //记忆化搜索时基于这样的事实--我们利用Dijkstra找到的
    44             dp[v]+=mDFS(i,n);                    //从2号点到1号点的最短路径中的每个点v,都有dist[v]<dist[1]。    
    45     return dp[v];
    46 }
    47 
    48 int main()
    49 {
    50     int n,m,a,b,d,i,j;
    51     while(scanf("%d",&n),n)
    52     {
    53         scanf("%d",&m);
    54         for(i=1;i<=n;++i)
    55             for(j=1;j<i;++j)
    56                 map[i][j]=map[j][i]=0x3fffffff;
    57         memset(dp,-1,sizeof(dp));
    58         for(i=1;i<=m;++i)
    59         {
    60             scanf("%d%d%d",&a,&b,&d);
    61             map[a][b]=map[b][a]=d;
    62         }
    63         for(i=1;i<=n;++i)
    64             dis[i]=map[2][i];
    65         dijkstra(2,n);
    66         printf("%d\n",mDFS(1,n));
    67     }
    68     return 0;
    69 }
    功不成,身已退
  • 相关阅读:
    POJ 3320 Jessica's Reading Problem
    引用参数和传值参数的区别
    IBM 的数据库Informix 常用代语法
    设计模式之原型模式
    UBoot中使用tftp下载文件出现错误TFTP error: 'Access violation' (2)的解决办法
    printf格式控制符的完整格式
    如何打印hostent结构体中的所有数据
    Informix 数据库客户端 dbvisualizer SQL Commander 乱码解决方案
    设计模式之模版方法模试
    nfs: server 192.168.37.200 not responding, still trying的解决办法
  • 原文地址:https://www.cnblogs.com/dongsheng/p/2604143.html
Copyright © 2011-2022 走看看