Constructing Roads In JGShining's Kingdom(最长上升子序列)

Constructing Roads In JGShining's Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16956    Accepted Submission(s): 4819

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.题意:            题目给出2n个城市,分别分布在道路两旁,一边为富裕的城市,另一边为贫穷的城市,经过时代的发展,现在要在富裕与贫穷的城市之间建立道路相连,问满足条件的最多道路数. (条件: 道路两两不相交,且是符合一一对应的条件);题解:           简单DP,就是一个最长上升子序列的题目,不过题目很坑,在最后输出那里还要考虑道路的条数是否大于1,要是大于1,则需要输出roads.被这个坑了好久,要以此为鉴,以后看题目要仔细认真,看清楚,理解清楚题目后在动手打码.AC代码:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <map>#include <vector>#include <cmath>#include <cctype>using namespace std;typedef long long ll;const int M = 5e5 + 100;const int INF = 0xffffff;int d[M];char s[] = "s ";struct Node{    int x,y;    bool operator < (const Node &a) const    {        return a.y > y;    }    void readin()    {        scanf("%d %d",&x,&y);    }}p[M];int main(){    freopen("in","r",stdin);    int n,cnt = 0;    while(~scanf("%d",&n))    {        int to = 0;        for(int i = 0; i < n; i++)            p[i].readin();        sort(p,p + n);        for(int i = 0; i < n; i++)        {            int u = lower_bound(d,d + to,p[i].x) - d;            if(u == to) d[to++] = p[i].x;            else d[u] = p[i].x;        }        printf("Case %d:\n",++cnt);        printf("My king, at most %d road%scan be built.\n\n",to,to > 1 ? s : " ");    }    return 0;}