15HD_OJ——FatMouse's Speed

/*
 * Copyright (c) 2014, 烟台大学计算机学院
 * All rights reserved.
 * 文件名称：test.cpp
 * 作    者：李晓凯
 * 完成日期：2015年 6 月 20 日
 * 版 本 号：v1.0
 *
 * 问题描述：
 * 输入描述：
 * 程序输出：

 */

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 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

Sample Output

44597

解析：

对体重从小到大排序，要求体重越小，速度越快，找一个最长递增子序列即可

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;struct node    //老鼠的属性{    int w;    //体重    int s;     //速度    int x;     //输入时的顺序号}mice[1005];int cmp(node a,node b)   //以体重为标准，将老鼠按体重递减排序{    return a.w<b.w;}int main(){    int n=0;    int i,j;    int much[1005],head[1005];    //分别记录第i只老鼠为止，子序列长度与前驱存储位置（mice数组中下标）    while(scanf("%d %d",&mice[n].w,&mice[n].s)!=EOF)        mice[n].x=n+1,n++;    sort(mice,mice+n+1,cmp);   //排序    head[0]=-1;   //终止标记    much[0]=1;    int max1=0,x;    //由第一只老鼠到当前检索位置为止，子序列的最大长度和其前驱位置在前驱记录数组head中的下标    for(i=1;i<=n;i++)    {        head[i]=-1;        much[i]=1;        for(j=0;j<i;j++)        {            if(mice[i].w>mice[j].w && mice[i].s<mice[j].s && much[j]+1>much[i])    //体重严格递增，速度严格递减，子序列更长            {                much[i]=much[j]+1;                head[i]=j;                if(much[i]>max1)   //前驱处理                {                    max1=much[i];                    x=i;                }                //cout<<"i="<<i<<"j="<<j<<endl;            }        }    }    //cout<<"X==="<<x<<endl;    int road[1005];    i=0;    while(1)   //处理路劲记录    {        //cout<<"X="<<mice [x].x<<endl;        road[i++]=mice[x].x;        //cout<<x<<" "<<head[x]<<endl;        x=head[x];        if(x==-1)           break;    }    printf("%d\n",max1);    for(i--;i>=0;i--)        printf("%d\n",road[i]);    return 0;}