公共子序列1007
来源:互联网 发布:平面画图软件 编辑:程序博客网 时间:2024/06/10 01:47
Constructing Roads In JGShining's Kingdom
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 13
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
Problem Description
JGShining's kingdom consists of 2n(n is no more than 500,000) small cities which are located in two parallel lines.
Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource.
With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they're unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don't wanna build a road with other poor ones, and rich ones also can't abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.
Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II.
The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities ... And so as the poor ones.
But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.
For example, the roads in Figure I are forbidden.
In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^
Half of these cities are rich in resource (we call them rich cities) while the others are short of resource (we call them poor cities). Each poor city is short of exactly one kind of resource and also each rich city is rich in exactly one kind of resource. You may assume no two poor cities are short of one same kind of resource and no two rich cities are rich in one same kind of resource.
With the development of industry, poor cities wanna import resource from rich ones. The roads existed are so small that they're unable to ensure the heavy trucks, so new roads should be built. The poor cities strongly BS each other, so are the rich ones. Poor cities don't wanna build a road with other poor ones, and rich ones also can't abide sharing an end of road with other rich ones. Because of economic benefit, any rich city will be willing to export resource to any poor one.
Rich citis marked from 1 to n are located in Line I and poor ones marked from 1 to n are located in Line II.
The location of Rich City 1 is on the left of all other cities, Rich City 2 is on the left of all other cities excluding Rich City 1, Rich City 3 is on the right of Rich City 1 and Rich City 2 but on the left of all other cities ... And so as the poor ones.
But as you know, two crossed roads may cause a lot of traffic accident so JGShining has established a law to forbid constructing crossed roads.
For example, the roads in Figure I are forbidden.
In order to build as many roads as possible, the young and handsome king of the kingdom - JGShining needs your help, please help him. ^_^
Input
Each test case will begin with a line containing an integer n(1 ≤ n ≤ 500,000). Then n lines follow. Each line contains two integers p and r which represents that Poor City p needs to import resources from Rich City r. Process to the end of file.
Output
For each test case, output the result in the form of sample.
You should tell JGShining what's the maximal number of road(s) can be built.
You should tell JGShining what's the maximal number of road(s) can be built.
Sample Input
21 22 131 22 33 1
Sample Output
Case 1:My king, at most 1 road can be built.Case 2:My king, at most 2 roads can be built.
思路:因为n为 500,000不可能采用二维数组dp来求最长公共子序列,所以就开两个数组,一个来记录富裕的城市,因为贫穷城市是递增顺序的,
所以只要如果前一个富裕城市比后一个城市小就加1,n为500000,所以得采用二分法才能不超时。
代码:
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <stdlib.h>#include <cmath>using namespace std;int a[500050];int dp[500050];int n;int lcs(){ dp[1]=a[1]; int sum=1; for(int i=2;i<=n;i++) { int b=1,c=sum; while(b<=c) { int mid=(b+c)/2; if(a[i]>dp[mid]) { b=mid+1; } else c=mid-1; } dp[b]=a[i]; if(b>sum) { sum=b; } } return sum;}int main(){ int z=1; int c,d; while(~scanf("%d",&n)) { int i,j; for(i=0;i<n;i++) { scanf("%d%d",&c,&d); a[c]=d; } memset(dp,0,sizeof(dp)); printf("Case %d:\n",z); int e=lcs(); if(e==1)printf("My king, at most 1 road can be built.\n\n"); else printf("My king, at most %d roads can be built.\n\n",e); z++; } return 0;}
阅读全文
0 0
- 公共子序列1007
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列...
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 最长公共子序列
- 代码复用
- python的eval、exec函数使用总结
- 怎么让优酷视频在网页中自动播放方法
- 百练_2694逆波兰表达式
- Codeforces Round #423 (Div. 2) E. DNA Evolution(树状数组)
- 公共子序列1007
- 79. Word Search
- 【算法概论习题】8.8解答
- Java基础学习总结(109)——Jdk动态代理和cglib动态代理总结
- Mac开启NTFS挂载功能
- Windows 10 将 Ubuntu作为它的APP
- LeetCode
- Java nio一个简单例子和画出之间通讯简图
- Android侧滑控件DrawableLayout以及NavigationView的使用