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hdu2476String painter (区间DP)

2024-07-13 01:55:17   219人阅读

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，求最少需要对a进行多少次操作，才能将a变成b。每次操作时将a中任意一段变成任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)


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