hdu 3183 A Magic Lamp （RMQ）

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

A Magic Lamp

Problem Description

Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question, and then the genie will realize one of her dreams.
The question is: give you an integer, you are allowed to delete exactly m digits. The left digits will form a new integer. You should make it minimum.
You are not allowed to change the order of the digits. Now can you help Kiki to realize her dream?

 

Input

There are several test cases.
Each test case will contain an integer you are given (which may at most contains 1000 digits.) and the integer m (if the integer contains n digits, m will not bigger then n). The given integer will not contain leading zero.

 

Output

For each case, output the minimum result you can get in one line.
If the result contains leading zero, ignore it.

 

Sample Input

   
   
   
   

    
    
    
    178543 4 
1000001 1
100001 2
12345 2
54321 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    13
1
0
123
321
   
   
   
   

rmq 问题：题意很简单，求一行数字中除去其中m个数字，使其组成最小的一个数字 使用rmq解题，设源数字长为n，那么除去m个数字后剩下的还剩n-m个数字，组成最小的数字。 （1）因为剩下n-m个数字，那么在1到m+1位置中最小的那个数字必是结果中的第一个数字i， （2）然后从这个数字i位置的下个位置i+1开始到m+2位置的数字中最小的那个数字必定是 结果中第二个数字，以此类推下去向后找。 （3）为了保证数字最小所以要保证高位最小还要保证数字长度够用~~

这道题虽是RMQ，但是由于不是很理解RMQ，所以就自己用循环的方法实现了0MS过

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;
typedef  long long LL;

#define N 1010
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

int main ()
{
    char str[N], s[N];
    int m;

    while (scanf ("%s %d", str, &m) != EOF)
    {
        int len = strlen (str), pos = 0;
        int k = len-m, j = 0;
        met (s, 0);

        while (k)
        {
            int minx = INF, Index = 0;

            for (int i=pos; i<=m; i++)
            {
                if (minx > str[i]-'0')
                {
                    minx = str[i]-'0';
                    Index = i;
                }
            }
            s[j++] = str[Index];
            pos = Index+1;
            k--;
            if (m < len-1) m++;
        }

        int flag = 0;
        k = 0;//记录位数，当已经达到m位的时候结束循环
        for (int i=0; i<j;)
        {
            while (s[i] == '0' && !flag)
                i++;//处理前导0的情况

            s[k++] = s[i];
            flag = 1;
            i++;
        }

        s[k] = '\0';
        if (s[0] == '\0') s[0] = '0', s[1] = '\0';//处理len==m的情况
        puts (s);
    }
    return 0;
}