hdu 3183 A Magic Lamp(RMQ)

题目链接：hdu 3183 A Magic Lamp

题目大意：给定一个字符串，然后最多删除K个。使得剩下的组成的数值最小。

解题思路：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，而且取的数最小，查询最小的操作用RMQ优化。

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

using namespace std;

const int maxn = 10005;

int N, M, d[maxn][20];
char s[maxn];

void rmq_init() {
    N = strlen(s);
    for (int i = 0; i < N; i++)
        d[i][0] = s[i];
    for (int k = 1; (1<<k) <= N; k++) {
        for (int i = 0; i < N; i++)
            d[i][k] = min(d[i][k-1], d[i+(1<<(k-1))][k-1]);
    }
}

int rmq_query(int l, int r) {
    int k = 0;
    while ((1<<(k+1)) <= r - l + 1) k++;
    return min(d[l][k], d[r-(1<<k)+1][k]);
}

int main () {
    while (scanf("%s%d", s, &M) == 2) {
        rmq_init();
        M = N - M;

        int mv = 0;
        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 == '0' && flag)
                continue;

            flag = false;
            printf("%c", c);
        }

        if (flag)
            printf("0");
        printf("
");
    }
    return 0;
}