求單調序列的nlogn演算法-HDU-1025

Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11464    Accepted Submission(s): 3283


Problem DescriptionJGShining'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. ^_^ 

InputEach 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. 

OutputFor 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.  

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.HintHuge input, scanf is recommended. 

 
最初寫的所 n^2複雜度的，逾時了。看了別人的代碼，才知道可以最佳化倒 nlogn由於此題的資料是1到n連續的，可以轉化為求最大單調序列。這裡用s[j] 儲存在 最大線路條數為 j 時，所串連的的城市的最小下限。更新s[] 時，保證它還是單調遞增的，用二分尋找最佳化。

#include <stdio.h>int s[500001];int a[500001];int main(){ int i,p,r,n,len,t=0;     while( ~scanf("%d",&n) )     {    for(i=0;i<n;i++)      {      scanf("%d%d",&p,&r);      a[p-1]=r;      }           s[0]=a[0];len=1; int start,end,mid;           for(i=1;i<n;i++)             {             //如果當前串連,可直接加入,增加總單調序列長度，並把當前長度的上限置為a[i]               if( a[i]>s[len-1] )  s[len++]=a[i];               else{//不可加入，則更新上限值. 保證s[] 所單調遞增的               //利用二分尋找               start = 0;               end = len-1;               while(start <= end){                   mid = (start + end) / 2;                   if(s[mid] > a[i]) end = mid-1;                   else start = mid+1;               }               s[start]=a[i];               }               for(int k=0; k<=len; k++)               printf("%d ",s[k]);               puts("\n");             }            printf("Case %d:\n",++t);            if(len==1) printf("My king, at most %d road can be built.\n\n",len);            else  printf("My king, at most %d roads can be built.\n\n",len);       }     return 0;}