Hdu 2476 String painter(区间dp)
2015-01-06 15:44
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题目链接
Total Submission(s): 1853 Accepted Submission(s): 820
[align=left]Problem Description[/align]
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?
[align=left]Input[/align]
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.
[align=left]Output[/align]
A single line contains one integer representing the answer.
[align=left]Sample Input[/align]
[align=left]Sample Output[/align]
题意:给两个长度相同的由小写字母组成的字符串A,B。一次操作,可以将A串的一个连续区间变成同一个字母,问把A串变成B串最少需要操作多少次?
题解:贪心的想,对于最优策略,每次操作的区间,只有包含关系,没有交叉的关系。也就是说对于原问题把区间[1,n]的串变成目标串的最少操作次数,假设我们的策略是操作区间[i,j],那么原问题就分割成子问题,求把区间[l,i-1]的串变成目标串的最少操作次数,求把区间[i,j]变成目标串的最少操作次数,求把区间[j+1,r]变成目标串的最少操作次数。也就是说一次操作可以,将原问题分割成几个小的子问题。而子问题是满足无后效性的。所以我们可以用动态规划解决。
用dp[l][r][0...25]表示对于区间[l,r] 该区间的字母全为a或b......z,把该状态变到变到目标状态的最小操作次数。
用dp[l][r][26] 表示对于区间[l,r] 该区间的字母为原A串中的字母,把该状态变到目标状态的最小操作次数。
对于每次操作,我们先解决该区间的第1个字母:
假设该区间的首字母和目标串的相同:
dp[l][r][x]=dp[l+1][r][x];
反之:
1,一次操作只将首字母变到目标状态:
dp[l][r][x]=dp[l+1][r][x]+1。
2,如果B[l]==B[i],我们一次操作可以把区间[l,i]变成字母B[l]:
dp[l][r][x]=dp[l+1][i-1][B[l]-'a]+dp[i+1][r][x]+1;
详情见代码:
String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1853 Accepted Submission(s): 820
[align=left]Problem Description[/align]
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?
[align=left]Input[/align]
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.
[align=left]Output[/align]
A single line contains one integer representing the answer.
[align=left]Sample Input[/align]
zzzzzfzzzzz abcdefedcba abababababab cdcdcdcdcdcd
[align=left]Sample Output[/align]
6 7
题意:给两个长度相同的由小写字母组成的字符串A,B。一次操作,可以将A串的一个连续区间变成同一个字母,问把A串变成B串最少需要操作多少次?
题解:贪心的想,对于最优策略,每次操作的区间,只有包含关系,没有交叉的关系。也就是说对于原问题把区间[1,n]的串变成目标串的最少操作次数,假设我们的策略是操作区间[i,j],那么原问题就分割成子问题,求把区间[l,i-1]的串变成目标串的最少操作次数,求把区间[i,j]变成目标串的最少操作次数,求把区间[j+1,r]变成目标串的最少操作次数。也就是说一次操作可以,将原问题分割成几个小的子问题。而子问题是满足无后效性的。所以我们可以用动态规划解决。
用dp[l][r][0...25]表示对于区间[l,r] 该区间的字母全为a或b......z,把该状态变到变到目标状态的最小操作次数。
用dp[l][r][26] 表示对于区间[l,r] 该区间的字母为原A串中的字母,把该状态变到目标状态的最小操作次数。
对于每次操作,我们先解决该区间的第1个字母:
假设该区间的首字母和目标串的相同:
dp[l][r][x]=dp[l+1][r][x];
反之:
1,一次操作只将首字母变到目标状态:
dp[l][r][x]=dp[l+1][r][x]+1。
2,如果B[l]==B[i],我们一次操作可以把区间[l,i]变成字母B[l]:
dp[l][r][x]=dp[l+1][i-1][B[l]-'a]+dp[i+1][r][x]+1;
详情见代码:
#include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> #include<string> #include<math.h> #define nn 110 #define inff 0x3fffffff #define eps 1e-8 typedef long long LL; const LL inf64=LL(inff)*inff; using namespace std; char a[nn],b[nn]; int dp[nn][nn][27]; bool check(int x,int y) { if(y==26) return a[x]==b[x]; return b[x]-'a'==y; } int dfs(int l,int r,int s) { if(dp[l][r][s]!=-1) return dp[l][r][s]; if(l>r) return 0; if(l==r) { if(check(l,s)) return dp[l][r][s]=0; return dp[l][r][s]=1; } dp[l][r][s]=inff; if(check(l,s)) return dp[l][r][s]=min(dp[l][r][s],dfs(l+1,r,s)); else dp[l][r][s]=min(dp[l][r][s],dfs(l+1,r,s)+1); int i; for(i=l+1;i<=r;i++) { if(b[l]==b[i]) { dp[l][r][s]=min(dp[l][r][s],dfs(l+1,i-1,b[l]-'a')+dfs(i+1,r,s)+1); } } return dp[l][r][s]; } int main() { int i; while(scanf("%s%s",a,b)!=EOF) { int len=strlen(a); memset(dp,-1,sizeof(dp)); printf("%d\n",dfs(0,len-1,26)); } return 0; }
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