PKU1631Bridging signals(最长不降子序列)

/*
Bridging signals
Time Limit: 1000MS        Memory Limit: 10000K
Total Submissions: 4649        Accepted: 2563

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

4
6
4
2
6
3
1
5
10
2
3
4
5
6
7
8
9
10
1
8
8
7
6
5
4
3
2
1
9
5
8
9
2
3
1
7
4
6

Sample Output

3
9
1
4

Source
Northwestern Europe 2003
*/
 
#include <iostream> using namespace std; int b[40000]; int main(void) { int cases,num,data,ans,left,right,middle; cin>>cases; while(cases > 0) { cases--; cin>>num; ans = 0; for(int i = 0;i < num;i++) { cin>>data; if(ans == 0) { b[1] = data; ans++; } else if(data > b[ans]) { ans++; b[ans] = data; } else { left = 0,right = ans; while(left < right) { middle = (left + right) / 2; if(b[middle] > data) { right = middle; } else { left = middle + 1; } } if(b[left] > data) { b[left] = data; } } } cout<<ans<<endl; } return 0;

 

 

终于见识到了传说中的最长不降子序列的问题，参考了别人的代码，更新data 的时候使用二分法不然会超时。

首先是得先看明白题目，这个题目的壳子装的很好，实际上就是最长不降子序列的问题（感叹自己给题目脱壳的能力实在太差了。。。），再说明白点就是完全不看题目就看题目给的输入再联想到最长不降子序列就焕然大悟了。。。。。

最长不降子序列的问题说白了还是DP问题中最关键的找到我们应该DP什么的问题，这题DP的关键还是一句话------b[i]表示长度为i的所有子序列中最后一个数字最小的那个数，我们DP的就是这个，当新的数比这个大时就加上当比他小的时候就用二分法查找更新某个b[i]的值。