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. ^_^
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
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.
给出n条边 看题中的图 每条边是连接穷与富的路 第一行为富的城市 第二行为穷的城市
路不能相交 问你最多能够建立几条路
输入边数n 之后的n行 为给出的n条边分别连接的富城市与穷城市
我们可以把穷城市的ip 赋给富城市 。
假设一个数组poor poor[i]=j表示富城市i与穷城市j相连 这时就把穷城市的ip赋给了富城市
因为城市的顺序是按编号大小排序的 所以只要我们求出最大递增子序列即可
借图更好理解 看第二个例子 最大递增子序列就是3 4 5
从穷城市的1号开始 判断poor[i]的最大递增子序列
poor[1]=3 poor[2]=2 poor[3]=4 poor[4]=5
最后要注意输出时 road是复数时 为roads
#include<cstdio>
#include<cstring>
using namespace std;
int poor[500001];
int len[500001];
int dis[500001];
int find(int a,int l,int r)
{
while(l<=r)
{
int mid=(l+r)/2;
if(a==dis[mid])
{
return mid;
}
if(dis[mid]>a)
{
r=mid-1;
}
else
{
l=mid+1;
}
}
return l;
}
int main()
{
int n,p,r,cut=0;;
while(scanf("%d",&n)!=EOF)
{
memset(len,0,sizeof(len));
memset(dis,0,sizeof(dis));
cut++;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&p,&r);
poor[p]=r;
dis[i]=500001;
}
len[1]=1;
int ans;
dis[1]=poor[1];
for(int i=2;i<=n;i++)
{
ans=find(poor[i],1,n);
dis[ans]=poor[i];
len[i]=ans;
}
int max=0;
for(int i=1;i<=n;i++)
{
if(max<len[i])
{
max=len[i];
}
}
printf("Case %d:
",cut);
if(max==1)
printf("My king, at most %d road can be built.
",max);
else
{
printf("My king, at most %d roads can be built.
",max);
}
}
return 0;
}