HDU 1160 FatMouse's Speed

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

/*
找到一个最多的老鼠序列，使得序列中的老鼠的体重满足递增，相应老鼠的速度满足递减。
先按体重递增进行sort排序，然后按照体重找到最长递减子序列。
用动态规划取得局部小序列的最优解，得到最后的最优解。

输入多组数据，每组包括老鼠的体重和速度。
输出这些数据中的最长上升子序列长度及路径。

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

output
4
4
5
9
7

*/

#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int MAXN = 1000;

struct Mice {
    int w,s,n;    //重量,速度,序号
}m[MAXN + 10];

bool cmp(Mice a,Mice b) {       //w相同按照w从小到大排序，不同则按照y从大到小排序
    return (a.w != b.w)?a.w < b.w:a.s > b.s;
}

int dp[MAXN + 10];    //dp[i]表示以第i个数据结尾的符合要求的子列长度
int p[MAXN + 10];     //记录对应的上一个数据
int ans[MAXN + 10];   //存放最终结果下标

int main(){
    int cnt = 1,i,j;
    while(cin >>m[cnt].w >>m[cnt].s){
        m[cnt].n = cnt++;
    }
    sort(m+1,m+cnt,cmp);

    int maxl=0;     //最长序列长度
    int maxi;       //最长序列的最后一个数下标
    memset(dp,1,sizeof(dp));        //初始化
    for(i = 1;i < cnt;i++){
        for(j = 1;j < i; j++)
            if(m[i].w > m[j].w&& m[i].s < m[j].s&& dp[j]+1 > dp[i]){    //每一次内循环都去局部序列最优。
                 dp[i]=dp[j]+1;
                 p[i]=j;    //记录i的上一个为j；
                 if(dp[i]>maxl){
                       maxi=i;
                       maxl=dp[i]; 
                 }
            }
    }
    i=0;
    while(maxi != 0){               //将下标从大到小存入ans，倒序存入
        ans[i++] = maxi;
        maxi = p[maxi];
    }
    cout <<i <<endl;
    while(i--){                     //倒序输出，即下标从小到大
        cout <<m[ans[i]].n <<endl;
    }
    return 0;
}