HDU 2476 区间dp

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1410    Accepted Submission(s): 606


[align=left]Problem Description[/align]
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?
 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
A single line contains one integer representing the answer.
 

[align=left]Sample Input[/align]

zzzzzfzzzzz
abcdefedcba
abababababab
cdcdcdcdcdcd

 

[align=left]Sample Output[/align]

6
7

 

[align=left]Source[/align]
2008 Asia Regional Chengdu

 

给定两个字符串，每一次可以把某一段刷为同一个字母，求最少刷多少次，可以把字符串1刷为字符串2。

区间dp，先对字符串2进行预处理，然后扫描得出答案。

代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014/3/12 0:09:36
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
char s1[105],s2[105];
int dp[105][105];//dp[i][j]为i~j的刷法
int ans[105],i,j,k,len;
int main()  {
while(~scanf("%s%s",s1,s2)){
len = strlen(s1);
memset(dp,0,sizeof(dp));
for(j = 0; j<len; j++)
for(i = j; i>=0; i--)//j为尾，i为头
{
dp[i][j] = dp[i+1][j]+1;//先每个单独刷
for(k = i+1; k<=j; k++)//i到j中间所有的刷法
if(s2[i]==s2[k])
dp[i][j] = min(dp[i][j],(dp[i+1][k]+dp[k+1][j]));//i与k相同，寻找i刷到k的最优方案
}
for(i = 0; i<len; i++)ans[i] = dp[0][i];//根据ans的定义先初始化
for(i = 0; i<len; i++){
if(s1[i] == s2[i])ans[i] = ans[i-1];//如果对应位置相等，这个位置可以不刷
else{
for(j = 0; j<i; j++)ans[i] = min(ans[i],ans[j]+dp[j+1][i]);//寻找j来分割区间得到最优解
}
}
printf("%d\n",ans[len-1]);
}

return 0;
}