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
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
题目大概:
有第一个串,变成第二个串,需要最少多少步操作,每次操作可以把一个任意区间涂成一样的颜色。
思路:
如果直接dp一下,那么会有在两种转移方式,当第一个串的第i个和第二个串的第i个,sum【i】=sum【i-1】.这个比较简单。
但是如果不相等那,如果是sum【i】=sum【i-1】+1;那显然是错误的,如果,刷第i段的时候,可以和前面的某一部分一起刷,那么就会多算一次。
那么怎么算那,可以直接暴力枚举一下,sum【i】=min(sum【i】,sum【j】+dp【j+1】【i】)j 属于(0--i)
这样每次不相等,都从前面枚举一遍。这也是一种套路题,很多题都是这么做。但这道题的最佳dp【j+1】【i】没有给出来。也没有从前面得到。所以只能自己算一下,由空白串到第二个串的最佳刷的次数是多少,用一个区间dp求一下,每到遇到两个相同的颜色,就可以优化一下,一起刷。一直优化到最后。
代码:
#include <bits/stdc++.h> using namespace std; const int maxn=110; const int INF=0x3f3f3f3f; char a[maxn]; char b[maxn]; int dp[maxn][maxn]; int sum[maxn]; int main() { while(~scanf("%s%s",a,b)) { memset(dp,0,sizeof(dp)); memset(sum,0,sizeof(sum)); int n=strlen(a); //for(int i=0;i<n;i++)dp[i][i]=1; for(int i=1;i<=n;i++) { for(int l=0;l<n-i+1;l++) { int r=l+i-1; dp[l][r]=dp[l+1][r]+1; for(int k=l+1;k<=r;k++) { if(b[l]==b[k]) dp[l][r]=min(dp[l+1][k]+dp[k+1][r],dp[l][r]); } } } for(int i=0;i<n;i++) { sum[i]=dp[0][i]; } for(int i=0;i<n;i++) { if(a[i]==b[i]) { sum[i]=sum[i-1]; } else { for(int j=0;j<=i;j++) { sum[i]=min(sum[j]+dp[j+1][i],sum[i]); } } } printf("%d\n",sum[n-1]); } return 0; }