poj1631 dp 最长上升子序列LIS

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这篇博客介绍了如何使用动态规划（dp）解决POJ1631问题，即找到一个数列的最长上升子序列。博主首先提出了常见的O(n^2)解决方案，但由于数据规模（n=40000）导致超时，然后详细讲解了优化后的O(n log n)算法，通过维护一个单调递增的dp数组，利用二分查找提高效率。代码示例展示了整个过程。

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如题：http://poj.org/problem?id=1631

Bridging signals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11636		Accepted: 6348

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

Sample Output

Source

Northwestern Europe 2003

思路：求最长上升子序列的dp。容易想到的一般解法dp[i]表示前i个的最长子序列的长度。dp[i]=max(dp[j])+1。其中满足a[j]<a[i].

复杂度为O(n*n);题目中n为40000.超时。于是看到一种O（nlogn）的解法。

dp[i]:表示所有的长度为i的上升子序列中的最后一个元素的最小值。

因此首先dp[i]所有的值赋值为INF。对于每一个a[i],在dp数组中寻找第一个》=a[i]的位置，将a[i]插入。这样复杂度为O(n*n),但是注意到dp数组肯定是单调递增。因此二分搜索每一个a[i].最终复杂度降为O（nlogn）.

注意：这里的序列最后求出来是严格递增，而不是非递减。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define MAXP 40005
#define INF 0x0fffffff

int a[MAXP];
int dp[MAXP];

int main()
{
// freopen("C:\\1.txt","r",stdin);
int n;
cin>>n;
while(n--)
{
  int p;
  cin>>p;
  int i;
  for(i=0;i<p;i++)
  {
   int x;
   cin>>x;
   a[i]=x;
   dp[i]=INF;
  }
  for(i=0;i<p;i++)
  {
   *lower_bound(dp,dp+p,a[i])=a[i];
  }
  int res=0;
  for(i=0;i<p;i++)
  {
   if(dp[i]==INF)
    break;
   res++;
  }
  cout<<res<<endl;
}
return 0;
}