String painter
Time Limit:3000MS Memory Limit:0KB 64bit
IO Format:%lld & %llu
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：给定两个长度相等的字符串，记为 A, B， 均只包含小写字母。每次可以把一个连续字串 “刷” 成相同的一个字母，问至少需要几次才能够把 A 变为 B。
思路：
假设一个空串要刷成目标串的极端情况（题意保证不存在空串，可以假设为所有字母均与目标串不同）。
用 dp[ i ][ j ] 表示把 A 串的 i-j 区间刷成目标串 B 的 i-j 所需要的最少步数，首先初始化所有的 dp[ i ][ i ] 为 1 。
dp[ i ][ j ] = dp[ i+1 ][ j ] + (strB[ i ] == strB[ j ]?0:1); 如果strB[ i ] == strB[ j ]，则 i，j 这两个位置可以同时刷。
同理，对于k=(i+1...j )如果str[k]==str[i] , 则dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]),,因为刷i的时候可以与k同时刷。
上面是对初始串与目标串完全不同的情况。
如果有部分的不同：
ans[ i ] 表示将 strA[ 1...i ] 刷成 strB [ 1...i ]的最小步数，
if strA[i]==strB[i], f[ i ] = f[ i-1 ];
else f[ i ] = min( f[ i ], f [ j ]+dp[ j+1 ][ i ]) ，1 <= j <= i-1
<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN = 110;
int dp[MAXN][MAXN];
int f[MAXN];
char strA[MAXN], strB[MAXN];
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int n;
while(scanf("%s %s", strA+1, strB+1) != EOF)
{
n = strlen(strA+1);
for(int i = 0; i <= n; i++)
dp[i][i] = 1;
for(int len = 2; len <= n; len++)
{ // 枚举区间长度
for(int i = 1, j = len; j <= n; i++, j++)
{
dp[i][j] = dp[i+1][j]+(strB[i]==strB[j]?0:1);
for(int k = i+1; k <= j-1; k++)
{
if(strB[i] == strB[k])
dp[i][j] = min(dp[i][j], dp[i+1][k]+dp[k+1][j]);
}
}
}
for(int i = 1; i <= n; i++)
{
f[i] = dp[1][i];
}
if(strA[1] == strB[1])
f[1] = 0;
for(int i = 1; i <= n; i++)
{
if(strA[i] == strB[i])
f[i] = f[i-1];
else
{
for(int j = 1; j <= i-1; j++)
{
f[i] = min(f[i], f[j]+dp[j+1][i]);
}
}
}
cout<<f[n]<<endl;
}
return 0;
}</span>