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 ;
}
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