No.28 - 牛客OJ- 2019校招真题-拼多多-选靓号

暴力+贪心：

枚举变为K次0～9，每次都最小费用，最小字典序

最小字典序，贪心思路，同费用下，先大值变小值，从前往后边，再小值变大值，从后往前边。

$注：$
string初始化 string s(S) 相当于赋值
string初始化 string s(L,‘9’) 相当于最大序

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<iostream>
#define MM(x) memset(x,0,sizeof(x))
#define INF 0x3f3f3f3f
using namespace std;

int N,K,L;
string S;
int cnt[15]; // 统计每个数字的数量
queue<pair<int,string> > q;

void solve(char key){
    int need = K - cnt[key-'0'];
    if(need <= 0){
        q.push(make_pair(0,S));
        return;
    }
    //cout<<key<<" need: "<<need<<endl;
    string s(S);
    char KeyUp = key + 1;
    char KeyDown = key - 1;
    int W = 0;
    //cout<<"now:"<<key<<endl;
    while(need){
        //cout<<KeyUp<<" "<<KeyDown<<endl;
        //cout<<"W:"<<W<<endl;
        //cout<<"s:"<<s<<endl;
        if(KeyUp <= '9'){
            int w = abs(KeyUp - key);
            for(int i=0; i<L && need;i++){
                if(s[i] == KeyUp){
                    s[i] = key;
                    W += w;
                    need --;
                }
            }
            if(need != 0) KeyUp++;
        }
        if(KeyDown >= '0'){
            int w = abs(KeyDown - key);
            for(int i=L-1;i>=0 && need ;i--){
                if(s[i] == KeyDown){
                    s[i] = key;
                    W += w;
                    need --;
                }
            }
            if(need != 0) KeyDown--;
        }
    }
    //cout<<"W:"<<W<<endl;
    //cout<<"s:"<<s<<endl;
    //cout<<"\n\n\n"<<endl;
    //cout<<W<<" "<<s<<endl;
    q.push(make_pair(W,s));
}

int main(){
    MM(cnt);
    cin>>N>>K>>S;
    L = S.length();
    for(int i=0;i<L;i++) cnt[S[i]-'0'] ++;
    //for(int i=0;i<10;i++) cout<<i<<" "<<cnt[i]<<endl;
    for(int i=0;i<10;i++){
        // 判断每个数字
        solve(i + '0');
    }
    string ansS(L,'9');
    int ansW = INF;
    while(!q.empty()){
        if(ansW > q.front().first){
            ansW = q.front().first;
            ansS = q.front().second;
        }else if(ansW == q.front().first && ansS > q.front().second){
            ansS = q.front().second;
        }
        q.pop();
    }
    cout<<ansW<<endl<<ansS<<endl;
    return 0;
}