hdoj-5903 Square Distance-优快云博客

本文探讨了一个算法问题，即如何找到与给定字符串具有特定汉明距离的字典序最小的平方字符串。通过动态规划方法解决了这一挑战，并提供了完整的代码实现。

Square Distance

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 245 Accepted Submission(s): 83

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

dp[i][j]表示第i个字符到n/2个字符以及i+n/2到n个字符与原字符串之间的距离是否能为j

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 100005
using namespace std;
typedef long long ll;

char str[1005];
int dp[505][1005];
int main(){
	
	//freopen("in.txt", "r", stdin);
	int t;
	
	scanf("%d", &t);
	while(t--){
		int n, m;
		scanf("%d%d%s", &n, &m, str+1);
		memset(dp, 0, sizeof(dp));
		dp[n/2+1][0] = 1;
		for(int i = n/2; i > 0; i--){
			if(str[i] == str[i+n/2]){
				for(int j = 0; j <= m; j++)
				 dp[i][j] = dp[i+1][j];
				for(int j = 0; j <= m-2; j++){
					if(dp[i+1][j])
					 dp[i][j+2] = 1;
				}
			}
			else{
				for(int j = 0; j <= m-1; j++)
				 dp[i][j+1] = dp[i+1][j];
				for(int j = 0; j <= m-2; j++)
				 if(dp[i+1][j])
				  dp[i][j+2] = 1;
			}
		}
		if(dp[1][m] == 0)
		 puts("Impossible");
		else{
			int k = m;
			for(int i = 1; i <= n/2; i++){
				for(int j = 0; j <= 25; j++){
					int p = 0;
					if(str[i] != j + 'a')
					  p++;
					if(str[i+n/2] != j + 'a')
					 p++;
			       
					if(dp[i+1][k-p]){
						str[i] = str[i+n/2] = j + 'a';
						k -= p;
						break;
					}
				}
			}
			puts(str+1);
		}
	}
	return 0;
}