zoukankan      html  css  js  c++  java
  • HDOJ 1160 FatMouse's Speed

    路径记录+LIS

    FatMouse's Speed

                                                              Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                          Total Submission(s): 6900    Accepted Submission(s): 3017
                                                                                                                                          Special Judge


    Problem Description
    FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.
     

    Input
    Input contains data for a bunch of mice, one mouse per line, terminated by end of file.

    The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.

    Two mice may have the same weight, the same speed, or even the same weight and speed. 
     

    Output
    Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that 

    W[m[1]] < W[m[2]] < ... < W[m[n]]

    and 

    S[m[1]] > S[m[2]] > ... > S[m[n]]

    In order for the answer to be correct, n should be as large as possible.
    All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one. 
     

    Sample Input
    6008 1300
    6000 2100
    500 2000
    1000 4000
    1100 3000
    6000 2000
    8000 1400
    6000 1200
    2000 1900
     

    Sample Output
    4
    4
    5
    9
    7
     

    Source
     

    Recommend
    Ignatius
     



    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>

    using namespace std;

    struct Fa
    {
        int id;
        int speed;
        int weight;
    }mouse[1100];

    int dp[1100];

    int fa[1100];

    bool cmp(Fa a,Fa b)
    {
        return a.weight<=b.weight;
    }

    int main()
    {
        int S,W;
        int cur=0;
    while(scanf("%d%d",&W,&S)!=EOF)
    {
        mouse[cur].speed=S;
        mouse[cur].weight=W;
        mouse[cur].id=cur;
        cur++;
    }

        sort(mouse,mouse+cur,cmp);

    /*
        for(int i=0;i<cur;i++)
            cout<<mouse.weight<<" , "<<mouse.speed<<endl;
      cout<<"___________________\n";
    */
        for(int i=0;i<=cur;i++)
           {
               dp=1;
               fa=i;
           }

        for(int i=1;i<cur;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(mouse.speed<mouse[j].speed&&dp<dp[j]+1&&mouse.weight>mouse[j].weight)
                   {
                       dp=dp[j]+1;
                       fa=j;
                   }
            }
        }
    /*
        for(int i=0;i<cur;i++)
            cout<<dp<<" ";
        cout<<endl<<endl;
        for(int i=0;i<cur;i++)
            cout<<fa<<" ";

        cout<<"\n----------------\n";
    */
       int p=-1,maxn=-1;
       for(int i=0;i<cur;i++)
       {
           if(dp>maxn)
           {
               maxn=dp;
               p=i;
           }
       }

        int ans[1100];
        int cnt=0;
        while(p!=fa[p])
        {
            ans[cnt]=mouse[p].id;
            p=fa[p];
            cnt++;
        }
        cout<<maxn<<endl;
        cout<<mouse[p].id+1<<endl;
        for(int i=cnt-1;i>=0;i--)
            cout<<ans+1<<endl;

        return 0;
    }


  • 相关阅读:
    圣杯+双飞翼 自适应布局
    drupal8 用户指南
    运维笔记:zabbix的运用(1)安装过程
    Nginx的初识
    PHP 获取LDAP服务器Schema数据
    Nginx(alias 和 root的区别)
    vue开发--生成token并保存到本地存储中
    PHP程序员必须知道的两种日志
    MVC 应用程序级别捕捉异常
    消息队列MQ
  • 原文地址:https://www.cnblogs.com/CKboss/p/3351041.html
Copyright © 2011-2022 走看看