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=500005;
int r[N],d[N];
int main()
{
    int n,cnt = 0,x,y,k;
    while(~scanf("%d",&n))
    {
        cnt++;
        memset(d,0,sizeof(d));
        for (int i = 0; i < n; i++)
        {
            scanf("%d %d",&x,&y);
            r[x] = y;
        }
        k = 1;
        d[k] = r[1];
        for (int i = 2; i <= n; i++)
        {
            int low = 1;
            int high = k;
            int mid;
            while(low <= high)
            {
                mid = (low+high)/2;
                if (d[mid] < r[i])
                    low = mid+1;
                else
                    high = mid-1;
            }
            d[low] = r[i];
            if (low > k)
                k = low;
        }
        if (k==1)
            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 0;
}