//dp[i]，长度为i的递增子序列的最后一位的最小值
//对于这种处理得到的dp[i]必然是递增的
//对于第i个数a[i]，只需要找到最小的大于它的位置pos，
//由于是最小的大于a[i] ,pos - 1 的值必然小于a[i]
//那么dp[pos] = a[i] ;
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
const int maxn = 500010 ;
int dp[maxn] ;
int a[maxn] ;
int find(int l , int r , int num)
{
while(l <= r)
{
int mid = (l + r) >> 1 ;
if(dp[mid] <= num)
l = mid + 1;
else r = mid - 1;
}
return l;
}
int main()
{
int n ;
int cas = 0;
while(~scanf("%d" ,&n))
{
for(int i = 1;i <= n;i++)
{
int t1 ,t2 ;
scanf("%d%d" ,&t1 ,&t2) ;
a[t1] = t2 ;
}
memset(dp , 0 ,sizeof(dp)) ;
int len = 1;dp[1] = a[1] ;
for(int i = 2;i <= n;i++)
{
int pos = find(1 , len ,a[i]) ;
dp[pos] = a[i] ;
if(pos > len) len ++ ;
}
printf("Case %d:\n",++cas);
if(len>1)
printf("My king, at most %d roads can be built.\n\n",len);
else
printf("My king, at most %d road can be built.\n\n",len);
}
return 0;
}
hdu 1025 dp+二分求最长递增子序列
原文：http://blog.csdn.net/cq_pf/article/details/45671349
