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hdu 2476 String painter

2024-07-10 23:01:07   225人阅读

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1529    Accepted Submission(s): 680


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
 

题意：
有两个字符串长度相同，现在有一个painter，一次可以把第一个字符串中的一段区间内的所有字母都换成同一个字母（这个字母可以是任意一个），问最少执行多少次操作，才能将第一个字符串s1转换成第二个字符串s2.

思路：
先将空白串变为s2，dp[i][j]-将i~j刷为目标串所需要的步数。
dp[i][j]=dp[i][j-1]+1;  j单独刷
当s[j]==s[k]时，j可以借助刷k的时候一起刷，dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);

处理完后，我们就要看s1需要喷刷多少次到s2了，s1比空白串的优点在哪里呢？在于s1有与s2相同的字符，这些字符可以不用个刷本身就与s2相同，用res[i]记录1~i区间第二个字符串得出的喷刷次数，如果第一个字符串的i位置与第二个字符串的i位置相同，那么这个位置就不用喷刷了，res[i]=res[i-1],如果不相同，就要就要借助一个位于1~(i-1)区间内的变量来分割开，res[i]=min(res[i],res[j]+dp[j+1][i])。

代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 105
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot;
char s1[maxn],s2[maxn];
int dp[maxn][maxn],res[maxn];

int main()
{
    int i,j,t,k,len;
    while(~scanf("%s%s",s1+1,s2+1))
    {
        memset(dp,0,sizeof(dp));
        n=strlen(s1+1);
        for(i=1;i<=n;i++)
        {
            dp[i][i]=1;
        }
        for(len=2;len<=n;len++)
        {
            for(i=1;i<=n;i++)
            {
                j=i+len-1;
                if(j>n) break ;
                dp[i][j]=dp[i][j-1]+1;
                for(k=i;k<j;k++)
                {
                    if(s2[k]==s2[j]) dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);
                }
            }
        }
        for(i=1;i<=n;i++)
        {
            res[i]=dp[1][i];
        }
        for(i=1;i<=n;i++)
        {
            if(s1[i]==s2[i]) res[i]=res[i-1];
            else
            {
                for(j=1;j<i;j++)
                {
                    res[i]=min(res[i],res[j]+dp[j+1][i]);
                }
            }
        }
        printf("%d\n",res[n]);
    }
    return 0;
}





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