CF 1367D Task On The Board

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题目：

Polycarp wrote on the board a string s containing only lowercase Latin letters ('a'-'z'). This string is known for you and given in the input.

After that, he erased some letters from the string s, and he rewrote the remaining letters in any order. As a result, he got some new string t. You have to find it with some additional information.

Suppose that the string t has length m and the characters are numbered from left to right from 1 to m. You are given a sequence of m integers: b1,b2,…,bm, where bi is the sum of the distances |i−j| from the index i to all such indices jj that tj>ti (consider that 'a'<'b'<...<'z'). In other words, to calculate bi, Polycarp finds all such indices j that the index j contains a letter that is later in the alphabet than ti and sums all the values |i−j|.

For example, if t = "abzb", then:

• since t1='a', all other indices contain letters which are later in the alphabet, that is: b1=|1−2|+|1−3|+|1−4|=1+2+3=6;
• since t2='b', only the index j=3 contains the letter, which is later in the alphabet, that is: b2=|2−3|=1;
• since t3='z', then there are no indexes j such that tj>ti, thus b3=0;
• since t4='b', only the index j=3 contains the letter, which is later in the alphabet, that is: b4=|4−3|=1.

Thus, if t = "abzb", then b=[6,1,0,1].

Given the string s and the array b, find any possible string t for which the following two requirements are fulfilled simultaneously:

• t is obtained from ss by erasing some letters (possibly zero) and then writing the rest in any order;
• the array, constructed from the string tt according to the rules above, equals to the array b specified in the input data.

思路：

我们可以确定的是位置上权值是0的字符，我们贪心的用第一个符合“0个数”的最大的字符去填充，然后我们把其他每一位减去这些最大字符提供的权值，这样我们又会找到第二大的字符，因为第二大的字符的权值是由第一大的字符提供的，按照这样执行，直到所有位置都被填充即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
 
using namespace std;
  
#define ll long long
#define pb push_back
#define fi first
#define se second
  
const int N = 2e6 + 10;
int cnt[N];
int v[N];
int vis[N];
char t[N];

void solve()
{
    int T;
    cin >> T;
    while(T--){

        //个数清空
        for(int i = 0; i <= 28; ++i) cnt[i] = 0;

        string s;
        cin >> s;
        //统计每个字符的个数
        for(auto ch : s){
            cnt[ch - 'a']++;
        }

        
        int n;
        cin >> n;
        //权值
        for(int i = 1; i <= n; ++i) cin >> v[i];
        //已经填过
        for(int i = 1; i <= n; ++i) vis[i] = false;

        vector<int > zero_inx;
        int len = 0;
        //使用到哪个字符
        int cnt_inx = 26;

        while(len != n){
            zero_inx.clear();
            for(int i = 1; i <= n; ++i){
                if(vis[i] || v[i]) continue;
                zero_inx.pb(i);
                vis[i] = true;
            }
            int _size = zero_inx.size(); 
            len += _size;
            char ch;
            for(int i = cnt_inx; i >= 0; --i){
                if(cnt[i] < _size) continue;
                ch = 'a' + i;
                cnt[i] = 0;
                cnt_inx = i;
                break;
            }

            //减去贡献的权值，填充字符
            for(int i = 0; i < _size; ++i){
                t[zero_inx[i]] = ch;
                int d = 0;
                for(int j = zero_inx[i] + 1; j <= n; ++j) v[j] -= ++d;
                d = 0;
                for(int j = zero_inx[i] - 1; j >= 1; --j) v[j] -= ++d;
            }
        }
        //cout << "====================" << endl;
        for(int i = 1; i <= n; ++i) cout << t[i];
        cout << endl;
        //cout << "====================" << endl;
    }

}
  
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    solve();
 
    return 0;
}