hdoj--1950--Bridging signals(二分查找+LIS)
Bridging signals
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1280 Accepted Submission(s): 832
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.
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.
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MIN -1000000
#define MAX 100001
int a[MAX];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(a,0,sizeof(a));
int n;
scanf("%d",&n);
int l,r,mid,top=0,m;
a[0]=MIN;
for(int i=0;i<n;i++)
{
scanf("%d",&m);
if(a[top]<m)
a[++top]=m;
else
{
l=1;r=top;
while(l<=r)
{
mid=(l+r)/2;
if(a[mid]<m)
l=mid+1;
else
r=mid-1;
}
a[l]=m;
}
}
printf("%d\n",top);
}
return 0;
}
hdoj--1950--Bridging signals(二分查找+LIS)的更多相关文章
- hdoj 1950 Bridging signals【二分求最大上升子序列长度】【LIS】
Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- HDU 1950 Bridging signals(LIS)
最长上升子序列(LIS)的典型变形,O(n^2)的动归会超时.LIS问题可以优化为nlogn的算法. 定义d[k]:长度为k的上升子序列的最末元素,若有多个长度为k的上升子序列,则记录最小的那个最末元 ...
- 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 ...
- hdu 1950 Bridging signals 求最长子序列 ( 二分模板 )
Bridging signals Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- (hdu)1950 Bridging signals(最长上升子序列)
Problem Description 'Oh no, they've done it again', cries the chief designer at the Waferland chip f ...
- HDU 1950 Bridging signals【最长上升序列】
解题思路:题目给出的描述就是一种求最长上升子序列的方法 将该列数an与其按升序排好序后的an'求出最长公共子序列就是最长上升子序列 但是这道题用这种方法是会超时的,用滚动数组优化也超时, 下面是网上找 ...
- 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 ...
- nyoj--214--单调递增子序列(二)(二分查找+LIS)
单调递增子序列(二) 时间限制:1000 ms | 内存限制:65535 KB 难度:4 描述 给定一整型数列{a1,a2...,an}(0<n<=100000),找出单调递增最长子序 ...
- 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, ...
随机推荐
- pyspark MLlib踩坑之model predict+rdd map zip,zip使用尤其注意啊啊啊!
Updated:use model broadcast, mappartition+flatmap,see: from pyspark import SparkContext import numpy ...
- C# 正则表达式
C# 正则表达式 正则表达式 是一种匹配输入文本的模式..Net 框架提供了允许这种匹配的正则表达式引擎.模式由一个或多个字符.运算符和结构组成. 定义正则表达式 下面列出了用于定义正则表达式的各种类 ...
- IOC-Castle Windsor映射
Castle最早在2003年诞生于Apache Avalon项目,目的是为了创建一个IOC(控制反转)框架.发展到现在已经有四个组件了,分别是ActiveRecord(ORM组件),Windsor(I ...
- Eclipse插件Lambok,实现自动生成Java代码
1.下载Lombok.jar http://projectlombok.googlecode.com/files/lombok.jar 2.运行Lombok.jar: java -jar D:\00 ...
- Codeforces 991E. Bus Number (DFS+排列组合)
解题思路 将每个数字出现的次数存在一个数组num[]中(与顺序无关). 将出现过的数字i从1到num[i]遍历.(i from 0 to 9) 得到要使用的数字次数数组a[]. 对于每一种a使用排列组 ...
- 读书笔记第三周 人月神话 刘鼎乾 PB16070837
读书笔记第三周:人月神话 这本书主要讲述了如何管理一个软件开发团队的问题,其中如何提高团队的效率可以说是本书的重点之一了.感觉这本书地中文版翻译得比较晦涩,很多表达比较模糊,看起来有些吃力,因此下 ...
- heavy dark--读《《暗时间》》
本书名为<<暗时间>>,个人觉得是一个非常好的名字:1.迷茫的大学生有多少的业余时间,但又浪费多少的业余时间,浪费的这莫多时间就如同人在黑夜中一样,大脑是在休息的状态.这是第一 ...
- iOS原生数据存储策略
一 @interface NSCache : NSObject Description A mutable collection you use to temporarily store transi ...
- Kattis - Babelfish
Babelfish You have just moved from Waterloo to a big city. The people here speak an incomprehensible ...
- BZOJ3529: [Sdoi2014]数表 莫比乌斯反演_树状数组
Code: #include <cstdio> #include <algorithm> #include <cstring> #define ll long lo ...