来吧，兄弟们，再来篇题解，其实也不是题解，是我自己的思路/心得/体会

Update Queries

题面翻译

题目翻译

一共 $t$ 组数据，每组数据给定长度为 $n$ 的字符串 $s$，长度为 $m$ 的 $ind$ 数组和字符串 $c$。

你可以任意安排 $ind$ 数组和字符串 $c$ 的顺序，并按照此顺序对字符串 $s$ 进行 $m$ 次操作：对于第 $i$ 次操作，将 $s_{ind_i}$ 的值更改为 $c_i$。要求经过 $m$ 次操作后的字符串 $s$ 的字典序最小，输出这个 $s$。

数据范围及约定

• $1 \le n,m \le 10^5$；
• 对于 $\forall i \in [1,m]$，有 $1 \le ind_i \le n$；
• $1 \le t \le 10^4$；
• $1 \le \sum n,\sum m \le 2 \times 10^5$；
• 保证字符串 $s$ 和字符串 $c$ 仅包含小写字母。

题目描述

Let's consider the following simple problem. You are given a string $ s $ of length $ n $ , consisting of lowercase Latin letters, as well as an array of indices $ ind $ of length $ m $ ( $ 1 \leq ind_i \leq n $ ) and a string $ c $ of length $ m $ , consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $ i $ -th operation, you set $ s_{ind_i} = c_i $ . Note that you perform all $ m $ operations from the first to the last.

Of course, if you change the order of indices in the array $ ind $ and/or the order of letters in the string $ c $ , you can get different results. Find the lexicographically smallest string $ s $ that can be obtained after $ m $ update operations, if you can rearrange the indices in the array $ ind $ and the letters in the string $ c $ as you like.

A string $ a $ is lexicographically less than a string $ b $ if and only if one of the following conditions is met:

• $ a $ is a prefix of $ b $ , but $ a \neq b $ ;
• in the first position where $ a $ and $ b $ differ, the symbol in string $ a $ is earlier in the alphabet than the corresponding symbol in string $ b $ .

输入格式

Each test consists of several sets of input data. The first line contains a single integer $ t $ ( $ 1 \leq t \leq 10^4 $ ) — the number of sets of input data. Then follows their description.

The first line of each set of input data contains two integers $ n $ and $ m $ ( $ 1 \leq n, m \leq 10^5 $ ) — the length of the string $ s $ and the number of updates.

The second line of each set of input data contains a string $ s $ of length $ n $ , consisting of lowercase Latin letters.

The third line of each set of input data contains $ m $ integers $ ind_1, ind_2, \ldots, ind_m $ ( $ 1 \leq ind_i \leq n $ ) — the array of indices $ ind $ .

The fourth line of each set of input data contains a string $ c $ of length $ m $ , consisting of lowercase Latin letters.

It is guaranteed that the sum of $ n $ over all sets of input data does not exceed $ 2 \cdot 10^5 $ . Similarly, the sum of $ m $ over all sets of input data does not exceed $ 2 \cdot 10^5 $ .

输出格式

For each set of input data, output the lexicographically smallest string $ s $ that can be obtained by rearranging the indices in the array $ ind $ and the letters in the string $ c $ as you like.

样例 #1

样例输入 #1

4
1 2
a
1 1
cb
4 4
meow
1 2 1 4
zcwz
7 4
abacaba
1 3 5 7
damn
7 10
traktor
7 6 5 4 3 2 1 6 4 2
codeforces

样例输出 #1

b
cwoz
abdcmbn
ccdeefo

提示

In the first set of input data, you can leave the array $ ind $ and the string $ c $ unchanged and simply perform all operations in that order.

In the second set of input data, you can set the array $ ind = [1, 1, 4, 2] $ and $ c = $ "zczw". Then the string $ s $ will change as follows: $ meow \rightarrow zeow \rightarrow ceow \rightarrow ceoz \rightarrow cwoz $ .

In the third set of input data, you can leave the array $ ind $ unchanged and set $ c = $ "admn". Then the string $ s $ will change as follows: $ abacaba \rightarrow abacaba \rightarrow abdcaba \rightarrow abdcmba \rightarrow abdcmbn $ .

我们解决这个题，首先要明白，什么是字典序。假设，从1到5,5个数。

最小情况就是1,2,3,4,5

最大就是5,4,3,2,1

序列相比是逐位对比的，从第一位开始，谁大谁的字典序就更大，如果一样就向后推，特殊的，类似这种，1比12的字典序要小。

结合输入分析题目，题目意思是给我们一个已知的序列s。我们要将ind数组结合字符串C按顺序修改字符串s并使其字典序最小。但我们可以任意将ind数组与c结合

如

4 4
meow
1 2 1 4
zcwz

原s数组是m e o w，c字符串是z c w z ，修改结果为c w o z。

如要让字符串的字典序最小，我们就需要让他按升序排列，在原序列的基础上对其进行修改，不仅是字符串需要升序，我们的修改也需要升序，这样的话，就可以最大限度的保证字典序最小。

特殊情况，例如这里面的1 1，有重复的，也就是说，这个值会被后来的值进行覆盖，我们可以利用这一点，如果有重复的修改点位，就将最大的（字典序）字母插入这里，后续被其他的所替代，代码上，我们可以直接删除重复的一步，并将最大字典序的字母删除。

所以该题的重点就是，去重以及排序.

代码如下#include

include<string.h>

include

include

using namespace std;
const int damn = 1e5 + 5;
char c[damn];
int num[damn];
char ind[damn];
int main()
{
int t;
cin >> t;
while (t--)
{
int n, m;
cin >> n >> m;
scanf("%s", c + 1);
for (int i = 1; i <= m; i++)
scanf("%d", &num[i]);
scanf("%s", ind + 1);
sort(ind + 1, ind + m + 1);
sort(num + 1, num + m + 1);
int cnt = 0;
for (int i = 1; i <= m; i++)
{
if (num[i] != num[i - 1])
{
c[num[i]] = ind[++cnt];
}
}
printf("%s\n", c + 1);
}
return 0;
}