HDOJ 3183 A Magic Lamp (贪心)

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2516    Accepted Submission(s): 984

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


    
    
    
     题意：给出一个字符串，删去m个数字，然后使得删除后的数字组成的数最小，没有前缀0 思路：刚看可能会直接对字符进行排序，然后删去最大的数字，但是想了想就会发现是错误的， 例如：21345 2 如果按照上面的方法，最后结果是213，但是正确的确实134，所以就不是 那么的简单了，写出来会发现一个规律，当s[i]>s[i+1]时，s[i]是要删除的，然后贪心每删除一个 就标记停止重新删除，如果符合条件的数字少于需要删除的数字，那么直接倒着删除就行了，因为 此时的字符串已经排好了序，直接删除相差的就行了。

ac代码：

    
    
    
    
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<iostream>
#include<algorithm>
#define fab(a) (a)>0?(a):(-a)
#define LL long long
#define MAXN 10010
#define INF 0xfffffff 
using namespace std;
char s[MAXN];
int v[MAXN];
int num[MAXN];
int main()
{
    int i,l,j;
    while(scanf("%s",s)!=EOF)
    {
        scanf("%d",&l);
        int len=strlen(s);
        for(i=0;i<len;i++)
        num[i]=s[i]-'0';
        memset(v,0,sizeof(v));
        int q=0;
        int k=l;
        while(k--)//贪心删除
        {
            for(i=0;i<len-1;i++)
            {
                if(!v[i])
                {
                    j=i+1;
                    while(v[j])//查找未标记的
                    j++;
                    if(num[i]>num[j])
                    {
                        v[i]=1,q++;//符合条件，标记停止
                        break;
                    }
                }
            }
        }
        int cnt=l-q;
        //printf("%d\n",cnt);
        for(i=len-1;i>=0;i--)//倒着寻找
        {
            if(!v[i]&&cnt)
            {
                v[i]=1;
                cnt--;
            }
        }
        int bz=0,o=0;
        for(i=0;i<len;i++)
        {
            if(!v[i])
            {
                if(s[i]!='0')
                bz=1;
                if(bz)
                printf("%c",s[i]),o++;
            }
        }
        if(o==0)
        printf("0");
        printf("\n");
    }
    return 0;
}