Convert the String (模拟)

 

 

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题意：给你两个字符串s1,s2;你需要用最小变化次数把是s1变成s2;
变化规则如下：

在s1中任意选择k个字符，选择k个字符中字典序最小的si,让k个字符都变成si;

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include <sstream>
#include<vector>
#include<cmath>    
#include<stack>
#include<time.h>
#include<ctime>
using namespace std;
#define inf 1<<30
#define eps 1e-7
#define LD long double
#define LL long long
#define maxn 100000005        
int ans[1001][1001] = {};
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int N, flag = 0, step = 0;
        scanf("%d", &N);
        string s1, s2;
        int Mina = 'z'-'a', Minb = 'z'-'a';
        cin >> s1 >> s2;
        for (int i = 0; i < N; i++)
        {
            if (s2[i] > s1[i])//如果s2比s1的字典序大，很显然是s1变不成的s2,s1只可能比原来小，不可能比原来大
            {
                flag = 1;//记上标记
                break;
            }
            Mina = min(Mina, s1[i] - 'a');//记录s1中字典序最小的字符
            Minb = min(Minb, s2[i] - 'a');//记录s2中字典序最小的字符
        }        
        if (Mina != Minb)//如果两个字符串最小的字符不同，则不可能变成相同的字符串
        {
            flag = 1;
        }
        for (int i = 0; i < N; i++)
        {
            int flag2 = 1;
            for (int j = 0; j < N; j++)
            {
                if (s2[i] == s1[j])//判断s2中的字符，s1是否都拥有，如果没有，s1也不可能变成s2
                {
                    flag2 = 0;
                    break;
                }
            }
            if (flag2 == 1)
            {
                flag = 1;
                break;
            }
        }
        if (flag == 1)//结束
        {
            printf("-1\n");
            continue;
        }
        while (s1 != s2)
        {
            int flag3 = 0;
            int step2 = 1;
            int Maxc;
            for (int i = 0; i < N; i++)
            {
                if (s1[i] > s2[i])//遍历s1,s2字符串，寻找s1大于s2的对应字符的s2字典序最大的,你不可能跳过最大的反而去变化比它小的
                {
                    if (flag3 == 0)
                    {
                        Maxc = s2[i]-'a';
                        flag3 = 1;
                    }
                    else
                    {
                        Maxc = max(Maxc, s2[i] - 'a');
                    }
                }
            }
            for (int i = 0; i < N; i++)
            {
                if (s1[i] - 'a' >= Maxc && s1[i] - 'a' >= s2[i] - 'a'&&s2[i]-'a'<=Maxc)//s1符合这三个条件
                {
                    ans[step][step2] = i;//记录下s1的位置
                    step2++;
                    s1[i] = Maxc + 'a';//把s1对应位置变成之前找到最大的s2字符
                }
            }
            ans[step][0] = step2 - 1;//有k个位置
            step++;
        }
        printf("%d\n", step);
        for (int i = 0; i < step; i++)
        {
            for (int j = 0; j <= ans[i][0]; j++)
            {
                if (j == 0)
                {
                    printf("%d", ans[i][j]);
                }
                else
                {
                    printf(" %d", ans[i][j]);
                }
            }
            printf("\n");
        }
    }
}