hdu1025 最大递增子序列的优化

Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12371 Accepted Submission(s): 3526

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. ^_^

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.

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.
Hint
Huge input, scanf is recommended.
 

Author

JGShining（极光炫影）

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/*此题数据量略大，所以用dp直接求果断TLE了，所以经这道题我学习了一下优化的算法，也就是在计算以当前项结尾的最长递增序列时使用二分查找，一次降低时间复杂度*///注意：输出时单数road复数要用roads，此细节wa了无数次。。。#include <iostream>#include <cstring>#include <cstdio>#include <deque>using namespace std;int r[500001];//b[i]存储的是长度为i的递增序列的最小末尾项int b[500001];deque<int> q;int DP(int n){    int i,left=1,mid=0,right=1,len=1;    b[1]=r[1];    for(i=2;i<=n;i++)//计算以每一项结尾的最长递增序列    {        right=len;        left=1;        while(left<=right)//二分查找        {            //对b数组，从下标的中值也就是长度开始查找，            //如果当前项比b[mid]大，            //那说明当前项可以插到长度比maxlen/2大的序列的尾部，            //于是将范围缩小到后半段，接着使用二分，            //反之亦然            mid=(left+right)/2;            if(b[mid]<r[i])                left=mid+1;            else                right=mid-1;        }        b[left]=r[i];        if(left>len)            len+=1;    }    return len;}int main(){    int n,t=0;    while(cin>>n)    {        t++;        memset(b,0,sizeof(dp));        int i,ans;        for(i=0;i<n;i++)        {            int rr,pp;            scanf("%d%d",&rr,&pp);            r[rr]=pp;        }        ans=DP(n);        cout<<"Case "<<t<<":"<<endl;        if(ans==1)        cout<<"My king, at most "<<ans<<" road can be built."<<endl;        else        cout<<"My king, at most "<<ans<<" roads can be built."<<endl;        cout<<endl;    }    return 0;}