hdu 2476 String painter

最新推荐文章于 2021-07-31 18:20:15 发布

原创最新推荐文章于 2021-07-31 18:20:15 发布 · 347 阅读

0 ·

CC 4.0 BY-SA版权

区间dp 专栏收录该内容

6 篇文章

订阅专栏

本文探讨了一道复杂的区间动态规划问题——如何将一个字符串通过最少步骤转换为另一个字符串，每步操作可以将任意区间的字符变为同一字符。文章提供了一个高效的算法解决方案，并附带详细代码实现。

String painter

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3694 Accepted Submission(s): 1719

Problem Description

There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other character you want. That is, after using the painter, the segment is made up of only one kind of character. Now your task is to change A to B using string painter. What’s the minimum number of operations?

Input

Input contains multiple cases. Each case consists of two lines:
The first line contains string A.
The second line contains string B.
The length of both strings will not be greater than 100.

Output

A single line contains one integer representing the answer.

Sample Input

  
   zzzzzfzzzzz
abcdefedcba
abababababab
cdcdcdcdcdcd

Sample Output

6
7

Source

2008 Asia Regional Chengdu

Recommend

lcy | We have carefully selected several similar problems for you: 2480 2481 2478 2482 2475

题意：给你两个字符串，要求从第一个字符串变到第二个字符串最少需要几步，每步可以把任意区间的字符串变成同一个任意字母的字符串。

思路：比较难的区间dp，首先区间dp算出从无变到目标字符串任意区间的最少步数，然后第二步再从头dp到尾，挺难解释的，看代码吧：

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<bitset>
#include <utility>
using namespace std;
#define maxn 105
#define inf 0x3f3f3f3f
typedef long long LL;
const int mod = 1e8;
char s1[maxn], s2[maxn];
int dp[maxn][maxn], ans[maxn];
int main(){
	while (~scanf("%s%s",s1,s2)){
		int len = strlen(s1);
		memset(dp, 0, sizeof(dp));
		for (int i = 0; i < len; i++)
			dp[i][i] = 1;
		for (int i = 0; i < len; i++){
			for (int j = 0; j + i < len; j++){
				dp[j][j + i] = dp[j + 1][j + i] + 1;
				for (int k = j + 1; k <= j + i; k++){
					if (s2[j] == s2[k])
						dp[j][j + i] = min(dp[j][j + i], dp[j + 1][k] + dp[k + 1][j + i]);
				}
			}
		}
		ans[0] = 0;
		for (int i = 0; i < len; i++){
			ans[i + 1] = dp[0][i];
			if (s1[i] == s2[i])
				ans[i + 1] = ans[i];
			else
				for (int j = 1; j <= i; j++)
					ans[i + 1] = min(ans[i + 1], ans[j] + dp[j][i]);
		}
		printf("%d\n", ans[len]);
	}
}