2018 ICPC沈阳站 热身赛B——CLS and LCS

Problem B. CLS and LCS

Claris is learning about the longest common subsequence problem.
(这里省略一段LCS的介绍，因为最长公共子序列是dp的经典问题，所以大家可以自行查询)
For homework she was given two strings A and B, both of length N and she had to determine the length of the LCS of A and B. She determined the answer to be K but lost B. Given A and K, help her find a possible value of B. It is possible that Claris may have made a mistake and no such B exists, in that case output “WRONGANSWER”(without quotes).

Input
The first line of input contains two integers N(1<= N<=2000), K(0<=K<=2000). The second line contains a string A consists of N lowercase letters.

Output
Output one line consisting of the string B of N lowercase letters, or “WRONGANSWER” if no B is valid. If there are multiple correct B output any of them.

Examples

standard input	standard output
3 2 cls	csl
3 4 cls	WRONGANSWER

Note
For the first example, another possible answer is “csy”.

• 题意 ： A和B两字符串的LCS长度是k，现给定A和k，输出一个B。
• 思路 ： 统计A中出现次数最少的字母，例如是a，就将B置为N个a，然后从前往后把A中不是a的字母替换B中字母，替换k-a个即可。
  例如 A ：acbdbaaad N = 9， K = 8.
  其中c个数最少，则B = ccccccccc
  K - c个数 = 7
  则B = acbdbaaac。
• 代码 ：

#include "bits/stdc++.h"
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define fori(i,l,u) for(int i = l;i < u;i++)
#define forj(j,l,u) for(int j = l;j < u;j++)
#define F first
#define S second
#define pb push_back
#define mk make_pair
typedef long long  ll;
typedef pair<int, int> pi;
typedef pair<string, int> ps;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<pi> vpi;
const int maxn = 1e5 + 6;
int n,k;
char a[maxn],b[maxn];
int res[maxn];
void init(){
    mem(res,0);
}
int main()
{
//    freopen("1.txt", "r", stdin);
    init();
    cin>>n>>k;
    fori(i, 0, n)cin>>a[i];
    fori(i, 0, n){
        res[a[i] - 'a'] ++;
    }
    int m = maxn;
    int p = 0;
    fori(i, 0, n){
        if (res[i] != 0) {
            if (m >= res[i]) {
                m = res[i];
                p = i;
            }
        }
    }
    char c = 'a' + p;
    fori(i, 0, n) b[i] = c;
    int num = 0;
    fori(i, 0, n){
        if (a[i] != c) {
            b[i] = a[i];
            num++;
            if (num == k-res[p]) {
                break;
            }
        }
    }
    if (num < k-res[p]) {
        cout<<"WRONGANSWER"<<endl;
    }
    else
    cout<<b<<endl;
    return 0;
}

• 遇到的问题 ： 一开始觉得自己技能点不够，不敢轻举妄动。这些构造类的题考验的就是思维，所以要仔细想。还是cf打得不够多啊。