CSU - 1270 Swap Digit (贪心)

最新推荐文章于 2023-03-04 11:45:22 发布

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本文介绍了一个算法问题，即通过有限次交换相邻数字的位置来构造出一个整数的最大可能形式。文章详细阐述了如何使用贪心策略从高位到低位逐位确定最大数字的方法，并提供了完整的C++实现代码。

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Description

Now we have a number, you can swap any two adjacent digits of it, but you can not swap more than K times. Then, what is the largest probable number that we can get after your swapping?
Input

There is an integer T (1 <= T <= 200) in the first line, means there are T test cases in total.

For each test case, there is an integer K (0 <= K < 106) in the first line, which has the same meaning as above. And the number is in the next line. It has at most 1000 digits, and will not start with 0.

There are at most 10 test cases that satisfy the number of digits is larger than 100.
Output

For each test case, you should print the largest probable number that we can get after your swapping.
Sample Input

3
2
1234
4
1234
1
4321

Sample Output

3124
4213
4321

思路：贪心

从高位到低位，把数字换成尽可能的最大的数字，直到交换的次数用完。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<set>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
#include<unordered_set>
#define Tcases int t; scanf("%d", &t)
using namespace std;
typedef long long LL;
const int maxn = 1000+10;
char num[maxn];
int main() {
#ifdef _DEBUG
    freopen("data.in", "r", stdin);
#endif // _DEBUG
    Tcases;
    int k;
    while (scanf("%d%s", &k,num) == 2) 
    {
        int len = strlen(num);
        for (int i = 0; i < len; i++) 
        {
            if (k <= 0) break;

            //找出num[i]后，k步内最大的数字
            char max = num[i]; 
            int id = i;
            for (int j = i + 1; j < len&&j <= i + k; j++) 
            {
                if (max < num[j]) 
                {
                    max = num[j];
                    id = j;
                }
            }

            //调整位置
            for (int j = id; j > i; j--) 
            {
                num[j] = num[j - 1];      
            }
            num[i] = max;
            k -= id - i;
        }
        printf("%s\n", num);
    }
    return 0;
}