FZU OJ 2111 Min Number (贪心）

时间：2015-05-23 00:04:53 阅读：149 评论：0 收藏：0 [点我收藏+]

Problem 2111 Min Number

Accept: 586 Submit: 1139
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Now you are given one non-negative integer n in 10-base notation, it will only contain digits (‘0‘-‘9‘). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].

For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.

Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?

Please note that in this problem, leading zero is not allowed!

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.

Output

For each test case, output the minimum number we can get after no more than M operations.

Sample Input

39012 09012 19012 2

Sample Output

901210921029

题意：给你一个数n，和交换次数m，要求经过m次交换之后的最小的数是多少，不能有前导零。

所以就把第一次的交换拿出来单独考虑，首先贪心查找非零的最小值，然后把最小值和第一位交换，每交换一次，交换次数m--。

后面的也是一样的，贪心查找之后位数的最小值，然后和前面的数进行比较，如果比前面的数小就交换。

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    char num[20000];
    int len, t, m, k, i, j;
    cin >> t;
    while (t--)
    {
        cin >> num >> m;
        len = strlen(num);
        if (m != 0)//m为1的情况
        {
            for (i=0, k=0; i<len; i++)
            {
                if (num[i] < num[k] && num[i] != '0')//比较最小值
                {
                    k = i;
                }
            }
            if (num[0] > num[k])
            {
                swap(num[0], num[k]);//交换
                m--;
            }
        }
        for (i=1; i<len && m != 0; i++)//从第二位开始找
        {
            for (j=i, k=j; j<len; j++)//查找最小值
            {
                if (num[j] < num[k])
                {
                    k = j;
                }
            }
            if (num[i] > num[k])
            {
                swap(num[i], num[k]);//交换
                m--;
            }
        }
        cout << num << endl;
    }
    return 0;
}

FZU OJ 2111 Min Number (贪心）

标签：贪心 acm 算法

原文地址：http://blog.csdn.net/whjkm/article/details/45921931

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