OpenJudge/Poj 1631 Bridging signals
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
6Sample Output
3
9
1
4Source
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
- POJ 1631 Bridging signals
Bridging signals Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 9441 Accepted: 5166 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- glusterfs repo
Installing Gluster For RPM based distributions, if you will be using InfiniBand, add the glusterfs R ...
- poj 2325 Persistent Numbers (贪心+高精度)
把输入数字每次从9-2除,能整除则记录该数字,最后从小到大输出. 应该算是水题,不过窝第一次写高精度除法,虽然1A,不过中间改了好多次. /****************************** ...
- nginx 多域名配置 (nginx如何绑定多个域名)
nginx绑定多个域名可又把多个域名规则写一个配置文件里,也可又分别建立多个域名配置文件,我一般为了管理方便,每个域名建一个文件,有些同类域名也可又写在一个总的配置文件里. 一.每个域名一个 ...
- 第一个php网页
<?php date_default_timezone_set('PRC'); if($_POST[ok]) { //echo "记录"; //echo "here ...
- hdoj 1495 非常可乐【bfs隐式图】
非常可乐 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...
- jQuery中get与eq的区别
get与eq的区别 .eq() 减少匹配元素的集合,根据index索引值,精确指定索引对象. .get() 通过检索匹配jQuery对象得到对应的DOM元素. 同样是返回元素,那么eq与get有什么区 ...
- C#- 控制台Timer
很少在控制台上用定时器,最近要用到,百度了一遍文章.很不错,摘下来,作备忘 关于C#中timer类 在C#里关于定时器类就有3个 1.定义在System.Windows.Forms里 2.定义在Sys ...
- Android_消息机制
Android通过Looper.Handler来实现消息循环机制. Android的消息循环是针对线程的,每个线程都可以有自己的消息队列和消息循环. Android系统中的Looper负责管理线程的消 ...
- 【转】Spring 4.x实现Restful web service
http://my.oschina.net/yuyidi/blog/352909 首先我们还是跟之前一样,创建一个maven项目,不过因为Spring Restful web service是基于Sp ...
- vbs运行批处理
dim wshellset wshell=createobject("wscript.shell") wshell.run "cmd /c sc query Spoole ...