zoukankan      html  css  js  c++  java
  • HDU 1160

    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. 

    InputInput 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. 
    OutputYour 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


    #include <bits/stdc++.h>
    #define ms(a) memset(a,-1,sizeof(a))
    using namespace std;
    
    
    const int M = 1009;
    
    int dp[M];
    int pre[M];
    
    struct node{
    	int w,v,id;
    	bool operator <(const node &n)const{
            if(w>=n.w)return true;
            else if (w==n.w) return v<n.v;
            else return false;
        }
    };
    node a[M];
    
    bool cmp(node a, node b){
    	return a.w < b.w;
    }
    
    
    int main()
    {
    	int i = 0,n = 0;
    	ms(pre);
    	while(scanf("%d%d",&a[i].w,&a[i].v) != EOF)
    	{
    		a[i].id = i+1;
    		i++;
    		n++;
    	}
    	sort(a,a+n);
    	for(int i=0;i<n;i++)dp[i]=1;
    	for(int i = 0; i < n; i ++)
    		for(int j = 0; j < i; j++)
    		{
    			if(a[i].w < a[j].w && a[i].v > a[j].v)
    			{
    				if(dp[i] < dp[j] + 1)
    				dp[i] = dp[j] + 1,pre[i] = j;
    			}	
    		}
    		
    	int ans=-1,pos = 0;
        for(int i=0;i<n;i++){
            if(ans<dp[i]){
                ans=dp[i];
                pos=i;
            }
        }
        printf("%d
    ",dp[pos]);
        while(pos!=-1){
            printf("%d
    ",a[pos].id);
            pos=pre[pos];
        }
    	return 0;
    }
    
  • 相关阅读:
    Oracle 与.NET Framework 数据类型映射
    mvc使用JsonResult返回Json数据(转)
    like参数替换
    jquery 等比缩放
    【linq to sql】已有打开的与此命令相关联的 DataReader,必须首先将它关闭
    脚本
    2012年计划——开始我的敏捷个人之行
    在Win7 64位机器上安装Oracle 10客户端以及PlSql
    词干提取算法Porter Stemming Algorithm解读
    开源搜索框架Lucene学习之分词器(2)——TokenFilter类及其子类
  • 原文地址:https://www.cnblogs.com/stul/p/10350495.html
Copyright © 2011-2022 走看看