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
zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd
Sample Output
67
我擦啊
为什么感觉区间DP的题意都很不好理解啊，看了一晚上，再加上别人代码的理解，终于理解题目想要我们做什么了
题意：
给出两个串s1和s2，一次只能将一个区间刷一次，问最少几次能让s1=s2
例如zzzzzfzzzzz，长度为11，，我们就将下标看做0~10
先将0~10刷一次，变成aaaaaaaaaaa
1~9刷一次，abbbbbbbbba
2~8:abcccccccba
3~7:abcdddddcba
4~6:abcdeeedcab
5:abcdefedcab
这样就6次，变成了s2串了
第二个样例也一样
0
先将0~10刷一次，变成ccccccccccb
1~9刷一次，cdddddddddcb
2~8:cdcccccccdcb
3~7:cdcdddddcdcb
4~6:cdcdcccdcdcb
5:cdcdcdcdcdcb
最后竟串尾未处理的刷一次
就变成了串2cdcdcdcdcdcd
所以一共7次
思路：这种球区间最优解的，明显就是区间DP了- -，需要注意的是，要将1串变为2串，可以说，主要是看2串两个相同字符之间的区间
#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\n",ans[len-1]);}return 0;}
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对于旅行，从来都记忆模糊。记不得都去了哪些地方，
