hdu 2476 (区间动态规划)

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5530    Accepted Submission(s): 2629

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

题解：先求空串转化为str2的值；考虑区间[i,j]，i<=k<=j，如果s[i]=s[k]，那么dp[i][j]=dp[i+1][k-1]+dp[k][j];即求区间[i,j]转化为区间[i+1,j]。如ac aba->c aba， 而 aba->ba。

由str1求str2的次数，记ans[i]为区间[0,i]转化的最优解，那么j=i+1,区间ans[j]=min(ans[j],ans[k]+dp[k+1][j])，0<=k<j;

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,n) for(int i=(a);i<(n);++i)
#define per(i,a,n) for(int i=(n-1);i>=(a);--i)
#define INF 1e9
const int mod=1e9+7;
#define N 110
char a[N],b[N];
int dp[N][N];
int main(){
    ios::sync_with_stdio(0);
    int n;
    while(scanf("%s",a)!=-1){
        scanf("%s",b);
        n=strlen(a);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;++i) dp[i][i]=1;
        for(int l=1;l<n;++l){
            for(int i=0;i+l<n;++i){
                int j=i+l;
                dp[i][j]=dp[i+1][j]+1;
                for(int k=i;k<=j;++k){
                    if(b[i]==b[k]){
                        dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);   //[i,j]等价于 [i+1,j] 
                    }                //aabaa ->abaa ->baa=2 ,aa->a=1;
                }
                //cout<<"l="<<l<<' '<<i<<' '<<j<<' '<<dp[i][j]<<endl;
            }
        }
        int ans[N];
        for(int i=0;i<n;++i){
            ans[i]=dp[0][i];   //全用空白串替换
            if(a[i]==b[i]){
                if(i==0) ans[i]=0;
                else ans[i]=ans[i-1];
            }
            for(int k=0;k<i;++k){
                ans[i]=min(ans[i],ans[k]+dp[k+1][i]);
            }
        }
        cout<<ans[n-1]<<endl;
    }
    return 0;
}