HDU 2476 String painter

String painter

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2476
64-bit integer IO format: %I64d      Java class name: Main

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

 

解题：区间dp

　　我们可以先假设最坏的情况下，把a刷成b，当然是当a是空白的时候 dp[i][j]表示把空串i到j刷成b的i到j需要最少多少步

这时候有dp[i][j] = min(dp[i][j],dp[i+1][k]+dp[k+1][j]) 当b[i] == b[k]的时候

 

为什么是dp[i+1][k]+dp[k+1][j]呢 因为刷k的时候 我们可以同时把i给刷掉在b[i] == b[k]的时候

 

现在的问题是a不是空白的，那么我们该如何处理

 

ans[i] = min(ans[i],ans[j]+dp[j+1][i])

 

把j+1到i这段 看成是空串刷过来的

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 120;
int dp[maxn][maxn],ans[maxn];
char a[maxn],b[maxn];
int main() {
    while(~scanf("%s%s",a,b)) {
        int n = strlen(a);
        memset(dp,0,sizeof dp);
        for(int i = n-1; i >= 0; --i)
            for(int j = i; j < n; ++j) {
                dp[i][j] = dp[i+1][j] + 1;
                for(int k = i+1; k <= j; ++k) {
                    if(b[i] == b[k])
                        dp[i][j] = min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
                }
            }
        memset(ans,0x3f,sizeof ans);
        for(int i = 0; i < n; ++i) {
            if(a[i] == b[i]) ans[i] = i?ans[i-1]:0;
            else ans[i] = dp[0][i];
            for(int j = i-1; j >= 0; --j)
                ans[i] = min(ans[i],ans[j]+dp[j+1][i]);
        }
        printf("%d
",ans[n-1]);
    }
    return 0;
}