HDU2476(区间dp)

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4905    Accepted Submission(s): 2316

Problem 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

 

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解题思路：这题比较巧妙，需要稍微进行转化，刚开始，令dp[i][j]表示把A区间i - j变成跟B串一样所需要的最小步数，但是，会发现，由于有A的限制，使得这个状态直接计算好像有点麻烦，所以我们进行一次转换，我们令dp[i][j]表示把一个空串变成跟B一样所需要的最小步数，这样就少了许多约束，可以直接转移计算，最后，我们利用这个dp值在A串上再做一次dp就行，令ans[i]，前i个字符变成跟B一样所需要的最小步数，然后转移很简单，直接枚举每个分段点就行。

#include <bits/stdc++.h>
using namespace std;
int inf = 0x3f3f3f3f;
const int maxn = 100 + 10;
char A[maxn];
char B[maxn];
int dp[maxn][maxn];
int ans[maxn];
void init()
{
    memset(dp, inf, sizeof(dp));
    memset(ans, inf, sizeof(ans));
}
int main()
{
    while(~scanf("%s%s", A + 1, B + 1))
    {
        init();
        int len = strlen(A + 1);
        for(int i = 1; i <= len; i++) dp[i][i] = 1;
        for(int i = 2; i <= len; i++)
        {
            for(int s = 1; s + i - 1 <= len; s++)
            {
                int t = s + i - 1;
                dp[s][t] = min(dp[s][t], dp[s + 1][t] + 1);
                for(int k = s + 1; k <= t; k++)
                {
                    int res;
                    if(k + 1 <= t) res = dp[k + 1][t];
                    else res = 0;
                    dp[s][t] = min(dp[s][t], dp[s][k] + res);
                    if(B[s] == B[k]) dp[s][t] = min(dp[s][t], dp[s + 1][k] + res);
                }
            }
        }
        ans[0] = 0;
        for(int i = 1; i <= len; i++)
        {

            if(A[i] == B[i]) ans[i] = ans[i - 1];
            else
            {
                ans[i] = min(ans[i], ans[i - 1] + 1);
                for(int j = i - 1; j >= 0; j--) ans[i] = min(ans[i], ans[j] + dp[j + 1][i]);
            }
        }
        printf("%d\n", ans[len]);
    }
    return 0;

}