[区间dp]Hdu2467 String painter

String painter

Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u
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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
2008 Asia Regional Chengdu

题目大意：给两个串，问串1变成串2最少需要多少次，改变可以连续一段变成一样的字母

有一种方法是：
直接用空白串来刷出s2的最小次数
dp[i][j]，则为刷出2串【i，j】的最小次数，，
dp[i][j]=dp[i+1][j]+1;
一个一个变
if(s2[i]==s2[k])
dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j])
当s2[i]==s2[k]时，就只需要刷i+1,k..刷k时顺便把i也一起了结了；
ans[i]初值是dp[1][i]
if（s1[i]==s2[i])ans[i]=ans[i-1]
ans[i]=min(ans[i],ans[k]+dp[k+1][j])

这里有两点自己写的时候疑问的地方：
1.循环事外层为什么是逆序循环；
2.为什么dp[i][j]的初始答案是从dp[i+1][j]推来的而不是dp[i][j+1](后一种会错）

具体解答在代码下面。

代码

#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif
#define INF 0x3f3f3f3f
#define clock CLOCKS_PER_SEC
#define cle(x) memset(x,0,sizeof(x))
#define maxcle(x) memset(x,127,sizeof(x))
#define mincle(x) memset(x,-1,sizeof(x))
#define minn(x1,x2,x3) min(x1,min(x2,x3))
#define cop(a,x) memcpy(x,a,sizeof(a))
#define FROP "hdu"
#define C(a,b) next_permutation(a,b)
#define LL long long
#define smin(x,tmp) x=min(x,tmp)
using namespace std;
const int N=105;
char s1[N],s2[N];
int dp[N][N];//把空白串变成1串的次数
int ans[N];//2串前i位变成1串的次数
int main()
{
    freopen(FROP".in","r",stdin);
    freopen(FROP".out","w",stdout);
    while(scanf("%s%s",s1+1,s2+1)!=EOF)
    {
        cle(dp);
        int len=strlen(s1+1);
        for(int i = len ; i >= 1; i--)
            for(int j = i ; j <= len ; j++)
            {
                dp[i][j]=dp[i+1][j]+1;
                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]);
            }
        ans[0]=0;
        for(int i = 1; i<= len; i++)
            ans[i]=dp[1][i];
        for(int i = 1; i <= len; i++)
        {
            if(s1[i]==s2[i])
            ans[i]=ans[i-1];
            for(int j = 1; j < i; j++)
                ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
        }
        printf("%d\n",ans[len]);
    }
    return 0;
}

1.dp时外层循环起点，逆序循环，因为dp[i][j]是从i+1,k+1层推来
次层终点顺序，，同理
内层枚举中间节点
2.第二点我也不知道啊，，，为什么，，为什么，往大神给我解答。。。QAQ，，感觉含义也差不多，，不过要WA