hdu 1160 排序+下降子序列变形 FatMouse's Speed

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

 

分析：这个题打眼一看，LIS变形题，但是之前看过的n*lgn 的怎么用都不对。而且，对于排完序后直接 变形 求解严格最大上升子序列的话，再对weight去重时会影响最优结果，所以这个题目，用最原始的n^2法，求解就行。

代码如下：

#include<cstdio>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h> 
using namespace std;

struct Node {
	int id,wei,speed;
}data[1005];
int dp[1010];
int pre[1010];

int cmp(Node a,Node b)
{
	return a.wei>b.wei;//反着排序，就可以直接用上升子序列的模板了，然后pre倒序输出，正好是需要的结果
}

int main()
{
	int cnt=0;
	while(~scanf("%d%d",&data[cnt].wei,&data[cnt].speed))
	{
		data[cnt].id=cnt+1;
		cnt++;
	}
	
	sort(data,data+cnt,cmp);
	memset(pre,-1,sizeof(pre));
	for(int i=0;i<cnt;i++)
		dp[i]=1;
	for(int i=0;i<cnt;i++)
	{
		for(int j=0;j<i;j++)
		{
			if(data[j].wei>data[i].wei && data[j].speed<data[i].speed && dp[i]<dp[j]+1)
			{
				dp[i]=dp[j]+1;
				pre[i]=j;//记录的是倒序的，后面倒着输出 
			}
		}
	}
	int ans=0;
	int p;
	for(int i=0;i<cnt;i++)
	{
		if(dp[i]>ans)
		{
			ans=dp[i];
			p=i;
		}		
	}
	printf("%d\n",dp[p]);
	while(p!=-1)
	{
		printf("%d\n",data[p].id);
		p=pre[p];
	}
	return 0;
}