HDU 5903 Square Distance (贪心+dp)-优快云博客

Square Distance

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 192 Accepted Submission(s): 61

Problem Description

A string is called a square string if it can be obtained by concatenating two copies of the same string. For example, "abab", "aa" are square strings, while "aaa", "abba" are not.

Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.

Peter has a string

s=s1s2...sn of even length. He wants to find a lexicographically smallest square string

t=t1t2...tn that the hamming distance between

s and

t is exact

m . In addition, both

s and

t should consist only of lowercase English letters.

Input

There are multiple test cases. The first line of input contains an integer

T , indicating the number of test cases. For each test case:

The first contains two integers

n and

(1≤n≤1000,0≤m≤n,n is even) -- the length of the string and the hamming distance. The second line contains the string

s .

Output

For each test case, if there is no such square string, output "Impossible" (without the quotes). Otherwise, output the lexicographically smallest square string.

Sample Input

  
   3
4 1
abcd
4 2
abcd
4 2
abab

Sample Output

  
   Impossible
abab
aaaa

【题意】给一个字符串t ，求与这个序列刚好有m个位置字符不同的由两个相同的串拼接起来的字符串 s，要求字典序最小的答案
【解题方法】参考： http://www.cnblogs.com/nicetomeetu/p/5905924.html
把字符串折半，分成0 - n/2-1 和 n/2 - n-1

　　dp[i][j] 表示第i位及之后的总代价为j可不可行

　　从第 n/2-1 位推回第 0 位，若dp[0][m] = 1，则存在

　　然后贪心对每一位从'a'试到'z'，选取接下来存在解的字符

【AC 代码】

//
//Created by just_sort 2016/9/26 9:58
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1005;
int dp[maxn][maxn];//dp[i][j]表示第i位及之后的总代价为j是否可行
char s[maxn];
int main()
{
    int T,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        scanf("%s",s);
        //INIT
        memset(dp,0,sizeof(dp));
        dp[n/2][0] = 1;
        //change state
        for(int i=n/2; i>=0; i--){
            if(s[i] == s[i+n/2]){
                for(int j = 0; j <= m; j++)   dp[i][j] = dp[i+1][j]; //no change
                for(int j = 0; j <= m-2; j++) if(dp[i+1][j]) dp[i][j+2] = 1; //change two
            }else{
                for(int j = 0; j <= m-1; j++) if(dp[i+1][j]) dp[i][j+1] = 1; //change one
                for(int j = 0; j <= m-2; j++) if(dp[i+1][j]) dp[i][j+2] = 1; //change two
            }
        }
        //get ans
        if(!dp[0][m]){
            puts("Impossible");
            continue ;
        }
        int k = m;
        for(int i = 0; i < n/2; i++)
        {
            for(int j = 0; j < 26; j++){
                int p = 0;
                if(s[i] != j+'a') p++;
                if(s[i+n/2] != j+'a') p++;
                if(dp[i+1][k-p]){
                    s[i] = j + 'a';
                    s[i+n/2] = j + 'a';
                    k = k - p;
                    break;
                }
            }
        }
        puts(s);
    }
    return 0;
}