HDU 1025.Constructing Roads In JGShining's Kingdom【最长上升子序列n×logn算法】【1月6】

Constructing Roads In JGShining's Kingdom

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

   
   
   
   

    
    
    
    2
1 2
2 1
3
1 2
2 3
3 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.

  
  
  
  

（上面输出有坑！road竟然有单复数！！！！！！害死人）

最长上升子序列的n×logn算法就是，来一个数，比末尾大就直接加入队尾，否则从队列中找到第一个比当前数大的用当前数将其替换，最后队列的长度就是解。

题意就是说一排富有的村子，一排贫穷的，然后给你n条路，不交叉，最多能修多少条。那可以按照富有（贫穷）的排序，然后找贫穷（富有）的村子序列的最长上升子序列就可以了，这样就能保证不相交，而且就是最优解。当然，因为数据大，n^2的算法是不可以的。代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 500000 + 10;
int f[maxn], g[maxn];
int n, p, r, k, le, ri, kase = 1;
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        k = 0;
        memset(g, 0, sizeof(g));
        for(int i = 0;i < n; ++i)
        {
            scanf("%d %d", &p, &r);
            f[p] = r;
        }
        for(int i = 1;i <= n; ++i)
        {
            if(f[i] > g[k]) g[++k] = f[i];//放入队列
            else//二分查找替换
            {
                le = 0;
                ri = k;
                while(le <= ri)
                {
                    int mid = (le+ri)/2;
                    if(g[mid] <= f[i]) le = mid + 1;
                    else ri = mid - 1;
                }
                g[le] = f[i];
            }
        }
        if(k == 1) printf("Case %d:\nMy king, at most 1 road can be built.\n\n", kase++);//被坑了
        else printf("Case %d:\nMy king, at most %d roads can be built.\n\n", kase++, k);
    }
    return 0;
}