Bellovin HDU 5748

Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: 

(a) The setup time for the first wooden stick is 1 minute. 
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup. 

You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2). 

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces. 

Output

The output should contain the minimum setup time in minutes, one per line. 

Sample Input

3 
5 
4 9 5 2 2 1 3 5 1 4 
3 
2 2 1 1 2 2 
3 
1 3 2 2 3 1

Sample Output

213

思路：最长上升子序列 通过本题学习了lower_bound   ps:调用lower_bound之前必须确定序列为有序序列，否则调用出错

函数lower_bound()在first和last中的前闭后开区间进行二分查找，返回大于或等于val的第一个元素位置。如果所有元素都小于val，则返回last的位置

代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f

using namespace std;

int a[100005];
int b [100005];
int dp[100005];
int main() 
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		for(int i = 1;i <= n;i++)
		{
			scanf("%d",&a[i]);
			dp[i] = 0;
			b[i] = INF;
		}
		for(int i = 1;i <= n;i++)
		{
			int x = lower_bound(b+1,b+n+1,a[i])-b;
			b[x] = min(b[x],a[i]);
			dp[i] = max(x,dp[i]);
		}
		for(int i = 1;i < n;i++)
		{
			printf("%d ",dp[i]);
		}
		printf("%d\n",dp[n]);
			
		
	}
	return 0;
}