题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2476
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
题意:
给定两个字符串,第一个是初始串,第二个是目标串,问你把初始串变到目标串最少需要多少串!
PS:
1:假设初始串是空串,然后进行区间dp,dp[i][j]表示区间[i, j]变到与目标串相同最少需要的步数,有:dp[i][j]=dp[i+1][j]+1;
2:如果s2[i] == s2[k],有:dp[i][j] = min(dp[i][j], dp[i+1][k]+dp[k+1][j])。
代码如下:
#include <cstdio> #include <cstring> #define maxn 117 int MIN(int a, int b) { if(a < b) return a; return b; } int main() { int dp[maxn][maxn];//区间[i, j]的最少次数 int a[maxn]; char s1[maxn], s2[maxn]; while(~scanf("%s",s1)) { scanf("%s",s2); int len = strlen(s1); memset(dp,0,sizeof(dp)); //求区间[i,j]最少次数; for(int j = 0; j < len; j++) { for(int i = j; i >= 0; i--) { dp[i][j] = dp[i+1][j]+1;//一个一个的刷 for(int k = i+1; k <= j; k++) { if(s2[i] == s2[k]) { dp[i][j] = MIN(dp[i][j], dp[i+1][k]+dp[k+1][j]);//最优解 } } } a[j] = dp[0][j]; } for(int i = 0; i < len; i++) { if(s1[i]==s2[i]) { a[i] = a[i-1]; //如果对应位置相等,可以不刷 continue; } for(int j = 0; j < i; j++) { a[i] = MIN(a[i],a[j]+dp[j+1][i]);//最优解 } } printf("%d\n",a[len-1]); } return 0; }
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