[2016-03-11][POJ][1631][Bridging signals]

[2016-03-11][POJ][1631][Bridging signals]

Bridging signals

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

Submit Status

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

• 时间:2016-03-11 14:32:42 星期五
  
• 题目编号:POJ 1631 Bridging signals
  
• 题目大意:有两块板,每块板有p个接口,左边的板的接口需要连接到右边的板的接口,问最多能有多少条线不相交
  
• 输入:
• cntcase数据组数
• 每组数据
• p接口数目
• 接下来p行,每行一个数字,第i行表示左边第i接口应该接到的右边接口的编号
• 输出:最大不相交的数目
  
• 分析:
  
• 首先看第一组数据的情况
• 1 2 3 4 5 6
• 4 2 6 3 1 5
• 如果左1连接了右4,那么右1 2 3就肯定不能连接了,因为肯定相交
• 如果连接了2 那么后面的3还是可以连接了,
• 所以,题目转换成了,给定一个序列,求最长上升子序列
  
• 方法:O(nlogn算法)
  
• 解题过程遇到问题:使用dp O(n^2)会T!!!  
  

#include <vector>

#include <list>

#include <map>

#include <set>

#include <deque>

#include <queue>

#include <stack>

#include <bitset>

#include <algorithm>

#include <functional>

#include <numeric>

#include <utility>

#include <sstream>

#include <iostream>

#include <iomanip>

#include <cstdio>

#include <cmath>

#include <cstdlib>

#include <cctype>

#include <string>

#include <cstring>

#include <cstdio>

#include <cmath>

#include <cstdlib>

#include <ctime>

using namespace std;

typedef long long LL;

#define CLR(x,y) memset((x),(y),sizeof((x)))

#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))

#define FORD(x,y,z) for(int (x)=(y);(x)>=(z);--(x))

#define FOR2(x,y,z) int (x);for((x)=(y);(x)<(z);++(x))

#define FORD2(x,y,z) int (x);for((x)=(y);(x)>=(z);--(x))

const int maxp = 40000 + 100;

int a[maxp],stk[maxp];

int main(){

        //freopen("in.txt","r",stdin);

        //freopen("out.txt","w",stdout);

        int cntcase;

        scanf("%d",&cntcase);

        while(cntcase--){

                int p,pcur = 0;

                scanf("%d",&p);

                FOR(i,0,p){

                        scanf("%d",a+i);

                }

                FOR(i,0,p){

                        if(!pcur){

                                stk[pcur++] = a[i];

                        }else {

                                if(a[i] > stk[pcur - 1]){

                                        stk[pcur++] = a[i];

                                }else {

                                        int pos = lower_bound(stk,stk+pcur,a[i]) - stk;

                                        stk[pos] = a[i];

                                }

                        }

                }

                printf("%d\n",pcur);

        }

        return 0;

}

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