交换一个数字的任意两个位置，指定K次的最值

最新推荐文章于 2023-11-18 10:59:42 发布

转载最新推荐文章于 2023-11-18 10:59:42 发布 · 1.4k 阅读

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原文链接：http://www.cnblogs.com/ccut-ry/p/9800059.html

文章标签：

#python #java

本文探讨了通过交换数字字符串中任意两个位置的数字来寻找K次交换后的最大和最小可能数值的问题。利用广度优先搜索(BFS)策略，文章提供了一种解决方法，详细介绍了如何在不引入前导零的情况下，通过贪心策略找到最优解。

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Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m−i. In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k

swaps?

InputThe first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.
OutputFor each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
Sample Input

5
12 1
213 2
998244353 1
998244353 2
998244353 3

Sample Output

12 21
123 321
298944353 998544323
238944359 998544332
233944859 998544332

题意：给你一个数字，可以交换任意两个位置，求交换K 次后的最大以及最小值

思路分析：比赛的时候写了个贪心，.... 各种试数据，试一组，WA一次，赛后写了个广搜，过了

代码示例：

int n, k;
char s[50], f[50];
int len;
struct node
{
    string str;
    int pos, num;    
};
queue<node>que;

int cal(string str) {
    int res = 0;
    for(int i = 0; i < len; i++) res = res*10+(str[i]-'0');
    return res;
}

void bfs1() {
    while(!que.empty()) que.pop();
    que.push({s+1, 0, 0});
    int ans = inf;
    
    while(!que.empty()){
        node v = que.front(); que.pop();
        int p = v.pos;
        if (v.num <= k) ans = min(ans, cal(v.str));
        if (p >= len) break;
        if (v.num > k) continue;
        int mmin = 1000;
        for(int i = p+1; i < len; i++) {
            if (p == 0 && v.str[i] == '0') continue;
            mmin = min(mmin, v.str[i]-'0');
        }
        if (mmin >= (v.str[p]-'0')){
            que.push({v.str, p+1, v.num});
            continue;
        }
        int sign = 0;
        for(int j = len-1; j > p; j--){
            sign = 1;
            if ((v.str[j]-'0') == mmin) {
                swap(v.str[j], v.str[p]);
                que.push({v.str, p+1, v.num+1});
                swap(v.str[j], v.str[p]);
            }
        }
        if (!sign) que.push({v.str, p+1, v.num});
    }
    cout << ans << " ";
}

void bfs2(){    
    //swap()
    while(!que.empty()) que.pop();
    que.push({s+1, 0, 0});
    int ans = -1;
    
    while(!que.empty()) {
        node v = que.front(); que.pop();
        int p = v.pos;
        if (v.num <= k) ans = max(ans, cal(v.str));
        if (p >= len) break;
        if (v.num > k) continue;
        int mmax = -1;
        for(int i = p+1; i < len; i++) mmax = max(mmax, v.str[i]-'0');
        if (mmax <= (v.str[p]-'0')) {
            que.push({v.str, p+1, v.num});
            continue;
        } 
        int sign = 0;
        for(int j = len-1; j > p; j--){
            sign = 1;
            if ((v.str[j]-'0') == mmax) {
                swap(v.str[j], v.str[p]);
                que.push({v.str, p+1, v.num+1});
                swap(v.str[j], v.str[p]);
            }
        }
        if (!sign) que.push({v.str, p+1, v.num});
    }
    cout << ans << endl;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t;
    
    cin >> t;
    while(t--){
        scanf("%s%d", s+1, &k);
        len = strlen(s+1);
        bfs1();
        bfs2();
    }
    return 0;
}

转载于:https://www.cnblogs.com/ccut-ry/p/9800059.html