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hdu 2476 String painter(区间dp)

2024-09-20 20:53:57   210人阅读

                                             String painter

                                                                                               Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                                      Total Submission(s): 4108    Accepted Submission(s): 1915
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
明天来解释 我想洗澡睡觉了
 1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cctype> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<stack> 9 #include<set>10 #include<vector>11 #include<algorithm>12 #include<string.h>13 typedef long long ll;14 typedef unsigned long long LL;15 using namespace std;16 const int INF=0x3f3f3f3f;17 const double eps=0.0000000001;18 const int N=1000+10;19 char a[N],b[N];20 int dp[N][N];21 int DP[N];22 int main(){23     while(gets(a)){24         gets(b);25         int len=strlen(a);26         memset(dp,INF,sizeof(dp));27         for(int i=0;i<len;i++)dp[i][i]=1;28         for(int t=1;t<len;t++){29             for(int i=0;i+t<len;i++){30                 int j=i+t;31                 if(b[i]==b[j])dp[i][j]=dp[i][j-1];32                 else{33                     for(int k=i;k<j;k++)34                     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);35                 }36             }37         }38         for(int i=0;i<len;i++){39             if(i==0&&a[i]==b[i])dp[0][i]=0;40             else if(a[i]==b[i])dp[0][i]=dp[0][i-1];41             for(int j=0;j<i;j++)42             dp[0][i]=min(dp[0][i],dp[0][j]+dp[j+1][i]);43         }44         cout<<dp[0][len-1]<<endl;45     }46 }

 

 

hdu 2476 String painter(区间dp)


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