(二分+dp) poj 1631 Bridging signals

Bridging signals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5919		Accepted: 3233

Description

'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up completely, making the signals on the chip connecting the ports of two functional blocks cross each other all over the place. At this late stage of the process, it is too expensive to redo the routing. Instead, the engineers have to bridge the signals, using the third dimension, so that no two signals cross. However, bridging is a complicated operation, and thus it is desirable to bridge as few signals as possible. The call for a computer program that finds the maximum number of signals which may be connected on the silicon surface without crossing each other, is imminent. Bearing in mind that there may be thousands of signal ports at the boundary of a functional block, the problem asks quite a lot of the programmer. Are you up to the task?

A typical situation is schematically depicted in figure 1. The ports of the two functional blocks are numbered from 1 to p, from top to bottom. The signal mapping is described by a permutation of the numbers 1 to p in the form of a list of p unique numbers in the range 1 to p, in which the i:th number specifies which port on the right side should be connected to the i:th port on the left side.Two signals cross if and only if the straight lines connecting the two ports of each pair do.

Input

On the first line of the input, there is a single positive integer n, telling the number of test scenarios to follow. Each test scenario begins with a line containing a single positive integer p < 40000, the number of ports on the two functional blocks. Then follow p lines, describing the signal mapping:On the i:th line is the port number of the block on the right side which should be connected to the i:th port of the block on the left side.

Output

For each test scenario, output one line containing the maximum number of signals which may be routed on the silicon surface without crossing each other.

Sample Input

4642631510234567891018876543219589231746

Sample Output

3914

/*
题目大意：求一条最长的上升子序列.题目解答：用一般的0(n^2)超时，只能采用0(nlogn).我们先来回顾一下poj 1887 Testing the CATCHER是怎么做的。
		：首先我们确定了题目具有最优子结构，然后我们开始递归构造最优子结构，定义一个数组opt[i]表示前i个字符包含第i个字符的最长下降序列,
		：然后又最优子结构递归出最优解公式：opt[i]=max(opt[j]+1),num[i] < num[j] && 0<=j<i<m。但是由于子结构的无序性，
		：你只能遍历一遍寻出当中最大的可行解，导致复杂度一下子上升到n^2那我们再想问什么我们递归建立的子结构是不具有递增性或者说无序的呢，
		：原因在于我们定义它的时候就确定了这个结果的产生（opt[i]前i个字符包含第i个字符的最长下降序列）要想要包含第i个字符那就不可能一定
		：保证opt[i]>opt[i-1]。例如：2 5 4 1 opt[3]=2; opt[4]=1;那么我们怎么创建一个有序的最优子结构呢？？要保证opt[i]>=opt[i-1]设opt[i]
		：为前i个字符的最长子序列，这个显然不行，我们缺少子解得信息，无法更新，也就不能进行递归求解.那么求解一个n个字符的串，我们需要什么
		：信息？？1：第i个字符（这个属于源数据）。2：前i个字符的最长子序列 3：前i个字符的最长子序列的最优一个字符到底是多少？，如果有多个
		：怎么办？先来看第三个条件的问题：如果有多个怎么办？例如 1 5 3 ....；前三个字符的最长子序列一看就知道是长度2，但是方案却又很多种
		：{1,5}和{1,3}找那一种呢？当然是选取其中最小的一个，可以让目标增长的空间最大化3~无穷大>5~无穷大。好了现在三个条件都有了，关键的
		：步骤来了，我们怎样存储2和3的两个条件opt[条件2]=[条件3]就是在前i个字符最长子序列.....(未完)

Source Code

Problem: 1631  User: wawadimu 
Memory: 608K  Time: 110MS 
Language: C++  Result: Accepted 

Source Code 
*/
#include<iostream>
using namespace std;

#define maxn 40010
int num[maxn];
int b[maxn];//b[k]=m ；长度为k的最长上升子序列中最小的数字值m
int opt[maxn];//opt[k]=l；前k个数字最长上升子序列的长度l
int p;//数组大小
int main()
{
	//freopen("1631.txt","r",stdin);
	int cas;
	int i,j,k;
	int l,r;
	int blen;//保存当前b数组的大小(即已经求得的最长的上升子序列)
	scanf("%d",&cas);
	while(cas--)
	{
		scanf("%d",&p);
		for(i=1;i<=p;i++)
			scanf("%d",&num[i]);
		b[1]=num[1];//只有一个数字是当然就是他本身
		blen=1;//长度
		for(i=1;i<=p;i++)//以此插入第i个数字
		{
			l=1; r=blen;
			while(l<=r)//寻找一个刚刚大于num[i]的b[k]
			{
				int m=(l+r)>>1;
				if(b[m] <num[i])
					l=m+1;
				else
					r=m-1;
			}
			b[l]=num[i];//将大数字替换成小数字
			opt[i]=l;//更新前i个数字的最长大小
			blen = blen < l ? l : blen;//更新最长子序列长度
		}
		printf("%d/n",blen);//输出答案
	}
	return 1;
}