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hdu2476String painter (区间DP)

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:给定两个字符串a和b,求最少须要对a进行多少次操作,才干将a变成b。每次操作时将a中随意一段变成随意一个字母所组成的段。
#include<stdio.h>
#include<string.h>
int min(int a,int b)
{
    return a>b?

b:a; } int main() { int dp[105][105],ans[105]; char a[105],b[105]; while(scanf("%s%s",a,b)>0) { int len=strlen(a); memset(dp,0,sizeof(dp)); for(int r=0;r<len;r++) for(int i=0;i<len-r;i++) { int j=i+r; dp[i][j]=dp[i+1][j]+1; for(int k=i+1;k<=j;k++) if(b[i]==b[k]) dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]); } for(int i=0;i<len; i++) ans[i]=dp[0][i]; for(int i=0;i<len; i++) if(a[i]==b[i]) { if(i==0)ans[i]=0; else ans[i]=ans[i-1]; } else{ for(int k=0;k<i;k++) ans[i]=min(ans[i],ans[k]+dp[k+1][i]); } printf("%d\n",ans[len-1]); } }



hdu2476String painter (区间DP)