【HDU2476】【String painter】

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2368    Accepted Submission(s): 1038


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


对于dp[i][j] 来说最差的就是1+dp[i+1][j]咯,那怎么减少呢。。。 如果(i,j]里面也有个k使得b[i] == b[k]那么得话, 可以刷[i,k]这么长的一段,那么就有可能减少1.  但是存在问题,如果有一段跨越了k怎么办。。。 嗯,这个问题是不会出现的, 因为在[i,k]刷了b[i].如果跨越了k那么之前在b[k]刷的颜色就没了,所以是不允许有跨段的, 所以要写成&&想写成 dp[i][j] = dp[i+1][k] + dp[k+1][j]。 为什么不可以跨段 可以思考  BRBR 。 然后后来的那个  一段 要么是可以分段刷的,要么是不可以分段刷的。-


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
    
char a[110];
char b[110];
int dp[110][110];
int ans[110];
int len;
int main()
{
	while(gets(a))
	{
		gets(b);
		memset(dp,0,sizeof(dp));
		len = strlen(b);
		for(int j=0;j=0;i--)
			{
				dp[i][j] = 1 + dp[i+1][j];
				for(int k=i+1;k<=j;k++)
				{
					if(b[i] == b[k])
					dp[i][j] = min(dp[i+1][k]+dp[k+1][j],dp[i][j]);
				}
			}
		}
		memset(ans,0,sizeof(ans));
		for(int i=0;i


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