hdu 2476 String painter 区间dp

String painter

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

Total Submission(s): 1482    Accepted Submission(s): 644

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

zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

 

Sample Output

67

 

Source

2008 Asia Regional Chengdu

 

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题意：给定两个相同长度的字符串A,B，每次操作可以使一个区间[l,r]变成同一个字符，问最少需要多少次操作时A变成B。

思路：

我们先考虑这样一个问题：假如A是空串，A最少需要多少次操作变成B。不难想到区间dp，设dp[i][j]为将A区间[i,j]染成B的最少操作数,那么dp[i][j]分为两部分；(1)dp[i][j]=min(dp[i][j],dp[i+1][j]+(B[i]==B[i+1]?0:1) )  (2)dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j])  其中k满足B[i]==B[k] 且 i<k<=j。

再考虑A串，设f[i]表示A[0]~A[i]==B[0]~B[i]的最小操作数，初始化时假设A与B没有一个字符对应相同，即f[i]=dp[0][i] , f[0]特殊处理,f[0]=A[0]==B[0]?0:1; 那么f[i]考虑两部分，(1) f[i]=min(f[i-1],f[i])  ,A[I]==B[i]   (2)  f[i]=min( f[i] ,f[j] +dp[j+1][i]) 其中0<=j<i ; 对(2)我们可以这样理解：将[j+1,r]视作空串。详见程序：

记忆化搜索：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=100+100;const int inf=0x3fffffff;int n;int ans[MAXN],dp[MAXN][MAXN];char s1[MAXN],s2[MAXN];void dfs(int i,int j){    if(j<=i)    {        dp[i][j]=j-i+1;        return;    }    if(dp[i][j]!=-1)        return ;    dfs(i+1,j);    dp[i][j]=dp[i+1][j]+(s2[i]==s2[i+1]?0:1);    for(int k=i+1;k<=j;k++)        if(s2[i]==s2[k])        {            dfs(i,k-1); dfs(k+1,j);            if(dp[i][k-1]+dp[k+1][j]<dp[i][j]||dp[i][j]==-1)                dp[i][j]=dp[i][k-1]+dp[k+1][j];        }}int main(){    //freopen("text.txt","r",stdin);    while(~scanf("%s%s",s1,s2))    {        int n=strlen(s2);        memset(dp,-1,sizeof(dp));        for(int i=0;i<n;i++)            dp[i][i]=1;        for(int i=0;i<n;i++)            for(int j=i;j<n;j++)                if(dp[i][j]==-1)            dfs(i,j);        for(int i=0;i<n;i++)        {            ans[i]=dp[0][i];            if(s1[i]==s2[i])            {                if(i==0)                    ans[i]=0;                else                    ans[i]=ans[i-1];            }            for(int j=0;j<i;j++)                ans[i]=min(ans[i],ans[j]+dp[j+1][i]);        }        printf("%d\n",ans[n-1]);    }    return 0;}

线性：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=100+100;int n;int ans[MAXN],dp[MAXN][MAXN];char s1[MAXN],s2[MAXN];int main(){    //freopen("text.txt","r",stdin);    while(~scanf("%s%s",s1,s2))    {        int n=strlen(s2);        memset(dp,0,sizeof(dp));        for(int i=0;i<n;i++)            for(int j=i;j<n;j++)                dp[i][j]=j-i+1;        for(int i=n-2;i>=0;i--)            for(int j=i+1;j<n;j++)            {                dp[i][j]=dp[i+1][j]+(s2[i]==s2[i+1]?0:1);                for(int k=i+1;k<=j;k++)                    if(s2[i]==s2[k])                        dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]);            }        for(int i=0;i<n;i++)            printf("%d ",dp[0][i]);        printf("\n");        for(int i=0;i<n;i++)        {            ans[i]=dp[0][i];            if(s1[i]==s2[i])            {                if(i==0)                    ans[i]=0;                else                    ans[i]=ans[i-1];            }            for(int j=0;j<i;j++)                ans[i]=min(ans[i],ans[j]+dp[j+1][i]);        }        printf("%d\n",ans[n-1]);    }    return 0;}