题意：给你一串字符串，一次可以修改一个或者多个连续的字符串，问最少操作多少次使得原串变成目标串
思路：一开始完全不知道怎么DP...首先我们假设一个空串，先求空串变成目标串的最少次数，令dp[i][j]为i到j的操作次数，那么显然有dp[i][j]=dp[i+1][j]+1，如果目标串有s[i]==s[k]，那么dp[i][j]=min(dp[i][j]，dp[i+1][k]+dp[k+1][j])
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 100000
#define LL long long
int cas=1,T;
int dp[105][105],a[105];
char s[105],t[105];
int main()
{
while (scanf("%s",s+1)!=EOF)
{
scanf("%s",t+1);
s[0]=t[0]=2;
int len = strlen(s)-1;
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
for (int i = 1;i<=len;i++)
dp[i][i]=1;
for (int i = len-1;i>=1;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 (t[i]==t[k])
dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
}
}
for (int i = 1;i<=len;i++)
{
a[i]=dp[1][i];
if (s[i]==t[i])
{
if (i==1)
a[i]=0;
else
a[i]=a[i-1];
}
else
for (int j = 1;j<i;j++)
a[i]=min(a[i],a[j]+dp[j+1][i]);
}
printf("%d\n",a[len]);
}
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
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
