hduacm 3183 rmq

http://acm.hdu.edu.cn/showproblem.php?pid=3183

问题等价与取N-M个数，每次取的时候保证后面能取的个数足够，并且取的数最小  查询最小用rmq

 #include<iostream>
 #include<algorithm>
 #include<cstdio>
 #include<cstring>

 using namespace std;

 const int maxn =  + ;

 int N,M,d[maxn][];
 char s[maxn];

 struct RMQ{
     void init()
     {
         N = strlen(s);
         for (int i = ;i<N;i++)
         d[i][] = s[i];
         for (int k = ;(<<k)<=N;k++)
         {
             for (int i = ;i<N;i++)
             d[i][k] = min(d[i][k-],d[i+(<<(k-))][k-]);

         }
     }
     int query(int L,int R)
     {
         int k = ;
         while (<<(k+)<=R-L+) k++;
         return min(d[L][k],d[R-(<<k)+][k]);
     }
 };

 RMQ rmq;

 int main()
 {
     while (~scanf("%s%d",s,&M))
     {
         rmq.init();
         M = N - M;
         int mv = ;
         bool flag = true;
         for (int i = M;i;i--)
         {
             int c = rmq.query(mv,N-i);
             while (mv<N&&s[mv]!=c) mv++;
             mv++;
             if (c==''&&flag)
             continue;
             flag = false;
             printf("%c",c);

         }
         if (flag)
         printf("");
         printf("\n");
     }
     return ;
 }