B - Bridging signals (LIS)
B - Bridging signals
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 rossing each other, is imminent. Bearing in mind that there may be housands 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?
Figure 1. To the left: The two blocks' ports and their signal mapping (4,2,6,3,1,5). To the right: At most three signals may be routed on the silicon surface without crossing each other. The dashed signals must be bridged.
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 pecifies 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.
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.
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
3
9
1
4
错误代码(一组数据行,怎么实现多组数据输入?)
#include <stdio.h>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f
using namespace std;
int dp[30020],a[30020];
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int m;
scanf("%d",&m);
for(int i=0; i<m ; i++)
{
scanf("%d",&a[i]);
dp[i]=INF;
}
for(int i=0; i<m; i++)
*lower_bound(dp,dp+n,a[i])=a[i]; //优化
printf("%d\n",lower_bound(dp,dp+m,INF)-dp);
}
return 0;
}
正确代码
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std; int a[41000];
int main()
{
int N,n,i,j,t;
scanf("%d",&N);
while(N--)
{
scanf("%d",&n);
scanf("%d",&t);
a[0]=t;
int top=1;
for(i=1; i<n; i++)
{
scanf("%d",&t);
if(t>=a[top-1])
a[top++]=t;
else
{
int z=0,q=top-1;
while(z<=q)
{
int mid=(z+q)/2;
if(a[mid]<t)
z=mid+1;
else
q=mid-1;
}
a[z]=t;
}
}
printf("%d\n",top);
}
return 0;
}
B - Bridging signals (LIS)的更多相关文章
- 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(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 ...
- HDU 1950 Bridging signals (LIS,O(nlogn))
题意: 给一个数字序列,要求找到LIS,输出其长度. 思路: 扫一遍+二分,复杂度O(nlogn),空间复杂度O(n). 具体方法:增加一个数组,用d[i]表示长度为 i 的递增子序列的最后一个元素, ...
- ZOJ 1093 Monkey and Banana (LIS)解题报告
ZOJ 1093 Monkey and Banana (LIS)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid= ...
- 浅谈最长上升子序列(LIS)
一.瞎扯的内容 给一个长度为n的序列,求它的最长上升子序列(LIS) 简单的dp n=read(); ;i<=n;i++) a[i]=read(); ;i<=n;i++) ;j<i; ...
- 最长递增子序列(LIS)(转)
最长递增子序列(LIS) 本博文转自作者:Yx.Ac 文章来源:勇幸|Thinking (http://www.ahathinking.com) --- 最长递增子序列又叫做最长上升子序列 ...
- Poj 2533 Longest Ordered Subsequence(LIS)
一.Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequenc ...
- Poj 3903 Stock Exchange(LIS)
一.Description The world financial crisis is quite a subject. Some people are more relaxed while othe ...
随机推荐
- LUA 删除元素的问题
table在删除元素时要注意,例t = { "hello", "world", "!"}t[1] = nil此时print(#t) --输出 ...
- UGUI 锚点设置为四方扩充模式然后设置局部坐标为0将出现什么问题
UGUI 锚点设置为四方扩充模式然后设置局部坐标为0将出现什么问题? 情形:按钮A挂在主画布上.四方扩充模式.A的中心和画面中心不重合. 这时候用代码设置A.localPosition = new V ...
- MVC4中压缩和合并js文件和样式文件
1.在App_Start文件夹中BundleConfig.cs类中添加相应的文件 1.1bundles.Add(new ScriptBundle("~/bundles/adminJs&quo ...
- Unpacking Argument Lists
[Unpacking Argument Lists] The reverse situation occurs when the arguments are already in a list or ...
- Golang之fmt格式“占位符”
golang的fmt包实现了格式化I/O函数: package main import "fmt" type Human struct { Name string } func m ...
- Laravel中Trait的用法实例详解
本文实例讲述了Laravel中Trait的用法.分享给大家供大家参考,具体如下: 看看PHP官方手册对Trait的定义: 自 PHP 5.4.0 起,PHP 实现了代码复用的一个方法,称为 trait ...
- [Jmeter]如何才能通过ant运行jmeter
在开始运行build.xml之前,还有一步必须要做,那就是将JMeter所在目录下extras子目录里的ant-JMeter-1.1.1.jar复制到Ant所在目录lib子目录之下,这样Ant运行时才 ...
- 阅读xtrabackup代码的一点笔记
xtrabackup binary最重要的两个过程是backup和prepare,对应的函数分别是xtrabackup_backup_func()和xtrabackup_prepare_func(), ...
- pyspider示例代码二:解析JSON数据
本系列文章主要记录和讲解pyspider的示例代码,希望能抛砖引玉.pyspider示例代码官方网站是http://demo.pyspider.org/.上面的示例代码太多,无从下手.因此本人找出一下 ...
- Access denied for user 'root'@'MiWiFi-Ryyy-srv' (using password: YES)
虽然是跟很多人一样的问题但是原因不同,其他很多文章说是授权问题,也确实是授权问题,但是,配置文件写的是连接localhost,而这里不知道什么原因切换了使用的用户,变成了默认访问MiWiFi-Ryyy ...