**【杭电1025】LIS 二分
来源:互联网 发布:json fromobject jar 编辑:程序博客网 时间:2024/04/29 01:59
Constructing Roads In JGShining's Kingdom
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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.两个for直接超时#include<cstdio>#include<algorithm>using namespace std;struct node{int x,y;}a[500001];int num[500001];int main(){int n;int k=1;while(~scanf("%lld",&n)){int maxm=0; for(int i=1;i<=n;i++) {scanf("%d%d",&a[i].x,&a[i].y);num[i]=1; for(int j=1;j<i;j++) { if(a[i].x>a[j].x&&a[i].y>a[j].y) if(num[i]<=num[j]+1)num[i]=num[j]+1;}maxm=max(maxm,num[i]);}printf("Case %d:\n",k++);if(maxm==1)printf("My king, at most %d road can be built.\n",maxm);elseprintf("My king, at most %d roads can be built.\n",maxm)}return 0;}
看了别人的,才知道要用二分。╭∩╮(︶︿︶)╭∩╮
详解:http://blog.csdn.net/lishuhuakai/article/details/8168328<span style="font-size:14px;">#include<iostream>using namespace std;long d[500001],a[500001]; //d[i]用来存最长不降子序列 最后一个元素 long find(long a,long left,long right) //二分查找{long mid;while(left<=right){mid=(left+right)/2; if(a>d[mid])//偏小 left=mid+1; else right=mid-1;}return left;}int main(){long n,i,j,len,p,t,k=0;while(scanf("%d",&n)!=EOF){k++;for(i=1;i<=n;i++){scanf("%d%d",&j,&p);a[j]=p;}d[1]=a[1];len=1;//初始序列长度 for(i=2;i<=n;i++) {t=find(a[i],1,len); //找出a[i]的位置d[t]=a[i]; //更新d[t],使d[t]尽量是最小的,显而易见,dp[t]越小,后面的dp[k](k>t)更大的可能性就会越大 //为什么可以这样做?仔细观察,发现对于一个a[i],要判断它是否是某个不降子序列的元素,我们只需要拿它和前面的dp[j](j<len)比较即可,//由于dp[j]记录的是到a[j]为止的序列的最长不降子序列的最后一个元素,如果a[i]>dp[j]并且a[i].b<dp[j+1],//那么我们自然要用a[i]更新dp[j+1](很明显,最长子序列长度为j+1),而如果a[i]比dp[j]的值都大,//这说明a[i]可以接在所有的dp[j]后面,自然我们选择最长的dp[len],因此,加入之后,len的长度要自加1 if(t>len) //因为d[len]都是最大的len++;} if(len==1) printf("Case %d:\nMy king, at most %d road can be built.\n\n",k,len); //单数 else printf("Case %d:\nMy king, at most %d roads can be built.\n\n",k,len); //复数}return 0;}</span>
0 0
- **【杭电1025】LIS 二分
- 杭电1025 LIS+排序
- POJ 1025(LIS+二分查找)
- 杭电-1950Bridging signals(LIS)
- 杭电5256 序列变换(LIS)
- 杭电1160 排序+LIS+记录
- 杭电 2817 二分
- 杭电二分题
- 杭电ACM1068(二分图)
- 【杭电1179】二分图
- 【杭电2063】二分图
- zoj 11986 LIS 二分
- poj3670,LiS(二分查找)
- LIS 二分加速版
- 合唱队形 LIS 二分
- LIS-二分查找模版
- 用二分做LIS
- 杭电-Super Jumping! Jumping! Jumping!(LIS)
- UVa 11093 Just Finish it up 模拟
- python之理解self传参
- 1032: 蛇行矩阵
- linux之tar批量解压文件
- 字符串转换成整数
- **【杭电1025】LIS 二分
- zzuli oj 1913 (模拟多个数相加)
- java23种设计模式九(连载)
- C# BitArray
- 1086: [SCOI2005]王室联邦
- 关于React Native 火热的话题,从入门到原理
- POJ 2342 (树状DP)入门
- 剑指Offer学习
- JSPatch