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
- 怎么加 一个 hyperlink 到 e-mail template for CRM
Recently I had a client inquire as to how one would insert a hyperlink into a CRM email template. Wh ...
- FPGA开发板
kingst.cnblogs.com 各种应用需要的接口不同: 做数字信号处理的,需要有AD/DA, 做图像处理,需要有图像接口.如果是通信的,需要有通信的接口,例如PCI/LVDS等.... 然后根 ...
- CM5(Cloudera Manager 5) + CDH5(Cloudera's Distribution Including Apache Hadoop 5)的安装详细文档
参考 :http://www.aboutyun.com/thread-9219-1-1.html Cloudera Manager5及CDH5在线(cloudera-manager-installer ...
- key_t键和ftok函数
系统建立IPC通讯(如消息队列.共享内存时)必须指定一个ID值.通常情况下,该id值通过ftok函数得到. ftok原型如下: key_t ftok( char * fname, int id ) f ...
- 动态添加DOM时,绑定的click事件会重复执行
最近因为业务需求,需要重写window的alert和confirm弹窗,但是每次显示的提示按钮不相同,所有每次打开的弹窗都需要重写生成,但是对于相同的按钮会保留上次创建时的click事件,所以当你创建 ...
- AMQP与QPID简介
国内私募机构九鼎控股打造APP,来就送 20元现金领取地址:http://jdb.jiudingcapital.com/phone.html内部邀请码:C8E245J (不写邀请码,没有现金送)国内私 ...
- JAVA中使用Redis
上节讲解了如何在centos上安装redis,点击查看.本节我们学习在java中使用redis.需要将jedis-*.jar添加到classpath(点击下载),如果使用连接池还需要commons-p ...
- libpcre.so.1 cannot be found
安装完Nginx之后,启动报错. [vagrant@localhost sbin]$ sudo ./nginx ./nginx: error while loading shared librarie ...
- listbox icon
. 实现过程 . . . 图 . 备注 . .关键点 . 相关链接 相关链接 相关链接 相关链接 相关链接. . 来自为知笔记(Wiz) 附件列表
- excel导入数据库iis设置
导入成功以后,基本这个小项目的所有功能都开发完成了,请IT部门帮我设定了一个固定IP,我以本机作为服务器,在本机IIS上发布了一个测试版,结果上传Excel数据报错, 错误信息“未在本地计算机上注册“ ...
