1.链接地址:

http://poj.org/problem?id=1631

http://bailian.openjudge.cn/practice/1631

2.题目:

Bridging signals
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 9882   Accepted: 5409

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

3.思路:

4.代码:

 #include "stdio.h"
//#include "stdlib.h"
#define NUM 40002
int dp[NUM];
int c[NUM];
int a[NUM];
int bsearch(int c[],int n,int a)
{
int l = ,r = n;
int m;
while(l<=r)
{
m=(l+r)/;
if(a>c[m] && a<=c[m+]) return m+;
else if(a<c[m]) r=m-;
else l=m+;
}
}
int LIS(int a[],int n)
{
int i,j,size=;
dp[]=;c[]=a[];
for(i=;i<=n;i++)
{
if(a[i] <= c[]) j=;
else if(a[i] > c[size]) j=++size;
else j= bsearch(c,size,a[i]);
c[j]=a[i];dp[j];
}
return size;
}
int main()
{
int n,p;
int i,j;
int ans;
scanf("%d",&n);
for(i=;i<n;i++)
{
scanf("%d",&p);
for(j=;j<=p;j++) scanf("%d",&a[j]);
int ans = LIS(a,p);
printf("%d\n",ans);
}
//system("pause");
return ;
}

OpenJudge/Poj 1631 Bridging signals的更多相关文章

  1. POJ 1631 Bridging signals

    Bridging signals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9441   Accepted: 5166 ...

  2. poj 1631 Bridging signals (二分||DP||最长递增子序列)

    Bridging signals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9234   Accepted: 5037 ...

  3. POJ 1631 Bridging signals(LIS O(nlogn)算法)

    Bridging signals Description 'Oh no, they've done it again', cries the chief designer at the Waferla ...

  4. POJ 1631 Bridging signals(LIS 二分法 高速方法)

    Language: Default Bridging signals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 1076 ...

  5. POJ 1631 Bridging signals & 2533 Longest Ordered Subsequence

    两个都是最长上升子序列,所以就放一起了 1631 因为长度为40000,所以要用O(nlogn)的算法,其实就是另用一个数组c来存储当前最长子序列每一位的最小值,然后二分查找当前值在其中的位置:如果当 ...

  6. POJ 1631 Bridging signals DP(最长上升子序列)

    最近一直在做<挑战程序设计竞赛>的练习题,感觉好多经典的题,都值得记录. 题意:给你t组数据,每组数组有n个数字,求每组的最长上升子序列的长度. 思路:由于n最大为40000,所以n*n的 ...

  7. POJ 1631 Bridging signals (LIS:最长上升子序列)

    题意:给你一个长为n(n<=40000)的整数序列, 要你求出该序列的最长上升子序列LIS. 思路:要求(nlogn)解法 令g[i]==x表示当前遍历到的长度为i的所有最长上升子序列中的最小序 ...

  8. Poj 1631 Bridging signals(二分+DP 解 LIS)

    题意:题目很难懂,题意很简单,求最长递增子序列LIS. 分析:本题的最大数据40000,多个case.用基础的O(N^2)动态规划求解是超时,采用O(n*log2n)的二分查找加速的改进型DP后AC了 ...

  9. POJ 1631 Bridging signals(LIS的等价表述)

    把左边固定,看右边,要求线不相交,编号满足单调性,其实是LIS的等价表述. (如果编号是乱的也可以把它有序化就像Uva 10635 Prince and Princess那样 O(nlogn) #in ...

随机推荐

  1. Codeforces335B - Palindrome(区间DP)

    题目大意 给定一个长度不超过5*10^4的只包含小写字母的字符串,要求你求它的回文子序列,如果存在长度为100的回文子序列,那么只要输出长度为一百的回文子序列即可,否则输出它的最长回文子序列 题解 这 ...

  2. 部署ActiveMQ的Share File System Master-Slave

    之前在项目里用MQ是用单节点,因为业务量不大没有主从.这样风险很大,会有单点问题.新项目起来了,需要一个高可用的MQ,故研究了下AMQ的几种master-slave方式: 1.Pure Master- ...

  3. windows 7 下安装 IIS 和 ArcGis Server 9.3 遇到的问题及解决方法

    windows 7 下安装 IIS 和 ArcGis Server 9.3 遇到的问题及解决方法 分类: ArcGIS server 计算机2012-07-31 14:17 631人阅读 评论(0)  ...

  4. JavaScript的递归之更多例子

    更多例子 第二个递归的例子是求两个自然数的最大公约数(有没有回到令人怀念的中学时代).下面的程序用的是经典的辗转相除法. //greatest common divisor //假定a.b都是正整数 ...

  5. 《Spring技术内幕》学习笔记17——Spring HTTP调用器实现远程调用

    1.Spring中,HTTPInvoker(HTTP调用器)是通过基于HTTP协议的分布式远程调用解决方案,和java RMI一样,HTTP调用器也需要使用java的对象序列化机制完成客户端和服务器端 ...

  6. [AngularJS + Webpack] require directives

    direictives/index.js: module.exports = function(ngModule) { //register all the directives here requi ...

  7. Linux 的启动流程-阮一峰

    http://www.ruanyifeng.com/blog/2013/08/linux_boot_process.html

  8. iOS开发中常见的语句@synthesize obj=obj的意义详解

    我们在进行iOS开发时,经常会在类的声明部分看见类似于@synthesize window=_window; 的语句,那么,这个window是什么,_ window又是什么,两个东西分别怎么用,这是一 ...

  9. QT事件

    qtevents多线程工作object存储 Another Look at Events(再谈Events) 最近在学习Qt事件处理的时候发现一篇很不错的文章,是2004年季刊的一篇文章,网上有这篇文 ...

  10. 详解ExplosionField的使用,实现View的粉碎效果

    小米平板卸载软件的时候,会有一个粉碎的效果,看起来很拉风,GitHub上有一个开源控件可以实现这个效果,我们一起来看看.先来看看效果图: 看起来不错吧,那我们今天就来详细说说ExplosionFiel ...