uva 10131 - Is Bigger Smarter?

Some people think that the bigger an elephant is, the smarter it is. To disprove this, you want to take the data on a collection of elephants and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the IQ’s are decreasing.

Input

The input will consist of data for a bunch of elephants, one elephant per line, terminated by the endof-file. The data for a particular elephant will consist of a pair of integers: the first representing its size in kilograms and the second representing its IQ in hundredths of IQ points. Both integers are between 1 and 10000. The data will contain information for at most 1000 elephants. Two elephants may have the same weight, the same IQ, or even the same weight and IQ.

Output

Say that the numbers on the i-th data line are W[i] and S[i]. 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 an elephant). If these n integers are a[1], a[2],..., a[n] then it must be the case that W[a[1]] < W[a[2]] < ... < W[a[n]] and S[a[1]] > S[a[2]] > ... > S[a[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 IQs 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

题意：找出一段不需按照原定顺序的子列使其满足条件

排序一次然后再最小上升（或下降）子序列

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n=1,m,w[1005],s[1005],dp[1005],lat[1005];
struct B{
	int w,s,id;
}a[1005];
bool cmp(B i,B j)
{
	if(i.s == j.s)
	  return i.w < j.w ;
	return i.s > j.s ;
}
void ready()
{
	ios::sync_with_stdio(false),cin.tie(0);
	while(cin>>a[n].w >>a[n].s)
	{
		a[n].id=n;
		n++;
	}
	n--;
	sort(a+1,a+n+1,cmp);
 
}
void work_for_w()
{
	dp[1]=1; 
	for(int i=2;i<=n;i++)
	{
		for(int j=1;j<=i;j++)
		{
			if(a[i].w>a[j].w && dp[j]+1>dp[i])
			{
				dp[i]=dp[j]+1;
				lat[i]=j;
			}
		}
	}
	int k=0,ans=0;
	for(int i=1;i<=n;i++)
	  if(dp[i]>dp[k])
	    k=i;
	while(k)
	{
		dp[++ans]=a[k].id;
		k=lat[k];
	}
	cout<<ans<<endl;
	for(int i=ans;i>=1;i--)
	  cout<<dp[i]<<endl;
}
int main()
{ 
	ready();
	work_for_w();
	return 0;
}