HDU2476 String painter(区间DP)

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1740    Accepted Submission(s): 770

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

题意：给定两个长度相等，只有小写字母组成的字符串s和t，每部可以吧s的一个连续子串“刷成同一个字母，问至少需要多少部才能把s 变成t；

        区间dp的经典题目；

 

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

char s1[105],s2[105];
int dp[105][105];//dp[i][j]为i~j的刷法
int ans[105],i,j,k,len;

int main()
{
    while(~scanf("%s%s",s1,s2))
    {
        len = strlen(s1);
        memset(dp,0,sizeof(dp));
        for(j = 0; j<len; j++)
        {
            for(i = j; i>=0; i--)//j为尾，i为头
            {
                dp[i][j] = dp[i+1][j]+1;//先每个单独刷
                for(k = i+1; k<=j; k++)//i到j中间所有的刷法
                {
                    if(s2[i]==s2[k])
                        dp[i][j] = min(dp[i][j],(dp[i+1][k]+dp[k+1][j]));//i与k相同，寻找i刷到k的最优方案
                }
            }
        }
        for(i = 0; i<len; i++)
            ans[i] = dp[0][i];//根据ans的定义先初始化
        for(i = 0; i<len; i++)
        {
            if(s1[i] == s2[i])
                ans[i] = ans[i-1];//如果对应位置相等，这个位置可以不刷
            else
            {
                for(j = 0; j<i; j++)
                    ans[i] = min(ans[i],ans[j]+dp[j+1][i]);//寻找j来分割区间得到最优解
            }
        }
          printf("%d
",ans[len-1]);
     }
    return 0;
}