Constructing Roads In JGShining's Kingdom(LIS)

http://acm.hdu.edu.cn/showproblem.php?pid=1025

题意：富人路与穷人路都分别有从1到n的n个点，现在要在富人点与穷人点之间修路，但是要求路不能交叉，问最多能修多少条。

思路：穷人路是按顺序给的，故求富人路的最长上升子序列即可。由于数据范围太大，应该用O(nlogn)的算法求LIS。

 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 using namespace std;
 const int N=;
 int r[N],d[N];
 int main()
 {
     int n,cnt = ,x,y,k;
     while(~scanf("%d",&n))
     {
         cnt++;
         memset(d,,sizeof(d));
         for (int i = ; i < n; i++)
         {
             scanf("%d %d",&x,&y);
             r[x] = y;
         }
         k = ;
         d[k] = r[];
         for (int i = ; i <= n; i++)
         {
             int low = ;
             int high = k;
             int mid;
             while(low <= high)
             {
                 mid = (low+high)/;
                 if (d[mid] < r[i])
                     low = mid+;
                 else
                     high = mid-;
             }
             d[low] = r[i];
             if (low > k)
                 k = low;
         }
         if (k==)
             printf("Case %d:\nMy king, at most %d road can be built.\n\n",cnt,k);
         else
             printf("Case %d:\nMy king, at most %d roads can be built.\n\n",cnt,k);
     }
     return ;
 }