hdu3183 RMQ

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2742    Accepted Submission(s): 1071

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

/*
hdu3183	RMQ
给你一串数字，求出删除m个数字后的最小值，用其他方法应该很好做的
开始想的是将里面的最大值选出来，但发现并不怎么靠谱，后来参考别人思路
发现可以通过RMQ求出最小值的位置，假设要在n个数中找出m个，那么第一个数一定在
[1,n-m]，找出这个数的位置x，那么第二个数则在[x+1,n-m+1]
hhh-2016-01-30 03:13:55
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <Map>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 1005;
char str[maxn];
char ans[maxn];
int dp[maxn][20];
int mm[maxn];

int Min(int i,int j)
{
    return str[i] <= str[j] ? i:j;
}

void iniRMQ(int n,char c[])
{
    for(int i = 0; i < n; i++)
    {
        dp[i][0] = i;
    }
    for(int j = 1; j <= mm[n]; j++)
    {
        for(int i = 0; i+(1<<j)-1 < n; i++)
            dp[i][j] = Min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
    }
}

int RMQ(int x,int y)
{
    int k = mm[y-x+1];
    return Min(dp[x][k],dp[y-(1<<k)+1][k]);
}

int main()
{
    int m;
    mm[0] = -1;
    for(int i = 1;i <= 1000;i++)
        mm[i] = ((i&(i-1)) == 0)? mm[i-1]+1:mm[i-1];
    while(scanf("%s%d",str,&m) != EOF)
    {
        int n = strlen(str);
        iniRMQ(n,str);
        int now,tot;
        now = tot = 0;
        m = n-m;
        while(m--)
        {
             now = RMQ(now,n-m-1);
            ans[tot++] = str[now++];
        }
        for(now = 0; now < tot; now++)
            if(ans[now] != '0')
                break;

        if(now == tot)
            printf("0\n");
        else
        {
            while(now < tot)
            {
                printf("%c",ans[now]);
                now ++;
            }
            printf("\n");
        }
    }
    return 0;
}