HDU 2476 String painter(区间dp)
2016-02-20 14:50
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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:给你两个字符串,从a串到b串,你每次可以选择将一个区间的全部字符变成一种字符求最小到达b串的次数。
思路:学习了kuangbin大神的博客才理解的,就是先dp获得一个从空串获得b串的最小次数,这个地方,比如获得abcda的代价就是代价(bcd)+1,这里的1就是涂后面那个a的代价,因为可以一起涂成a,然后涂中间的,所以这里直接b[i] == b[k]的时候,区间dp一下。然后对a串到b串,如果a[i] == b[i],这个i位置我我们就可以不涂,这里也区间dp一下。
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:给你两个字符串,从a串到b串,你每次可以选择将一个区间的全部字符变成一种字符求最小到达b串的次数。
思路:学习了kuangbin大神的博客才理解的,就是先dp获得一个从空串获得b串的最小次数,这个地方,比如获得abcda的代价就是代价(bcd)+1,这里的1就是涂后面那个a的代价,因为可以一起涂成a,然后涂中间的,所以这里直接b[i] == b[k]的时候,区间dp一下。然后对a串到b串,如果a[i] == b[i],这个i位置我我们就可以不涂,这里也区间dp一下。
#include<iostream> #include<cstdio> #include<algorithm> #include<map> #include<cstring> #include<map> #include<set> #include<queue> #include<stack> #include<cmath> using namespace std; const double eps = 1e-10; const int inf = 0x7fffffff, N = 1e2 + 10; typedef long long ll; using namespace std; char s , x ; int dp , ans ; int main() { while(~scanf("%s%s", s, x)) { int len = strlen(s); memset(dp, 0, sizeof(dp)); for(int i = 0; i<len; i++) for(int j = i; j<len; j++) dp[i][j] = j - i + 1; for(int i = len - 1; i>= 0; i--) for(int j = i + 1; j<len; j++) { dp[i][j] = dp[i+1][j] + 1; for(int k = i + 1; k<=j; k++) if(x[i] == x[k]) { dp[i][j] = min(dp[i][j], dp[i+1][k-1]+dp[k][j]); } } for(int i = 0; i<len; i++) { ans[i] = dp[0][i]; if(s[i] == x[i]) { if(!i) ans[i] = 0; else ans[i] = ans[i-1]; } for(int j = 0; j<i; j++) ans[i] = min(ans[i], ans[j] + dp[j+1][i]); } cout<<ans[len-1]<<endl; } return 0; }
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