hdoj 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): 2624    Accepted Submission(s): 1188

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

 

题意：给定两个字符串a和b，你可以选择在a串的任意一段涂上相同的字符，问你最少需要涂多少次才可以把a串变为b串。

思路：先计算出空串涂成b串的代价，用dp[i][j]存储区间[i, j]涂成和b串一样的最小代价。

因为字符都不相同，那么代价的优化点就在于b串中相同的字符。

dp[i][j] = min(dp[i+1][j]+1, dp[i+1][k]+dp[k+1][j]); (i+1<=k<=j && b[k] == b[i])

b[i]可以在涂区间[i+1, k]时顺带涂上。  处理时需要注意边界。

整理我们已经有的信息dp[i][j]，把空串变成b[i] - b[j]的字符串所需的最小代价。

设置ans[i]为a[0] - a[i]的字符串变成b[0] - b[i]所需的最小代价。

则状态转移：

a[i] == b[i]   ans[i] = ans[i-1];

a[i] != b[i]   ans[i] = min(ans[i], ans[j] + dp[j+1][i]); 把区间[j+1, i]当做空串来处理。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (100+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)using namespace std;char s1[110], s2[110];int dp[110][110], ans[110];int main(){    while(scanf("%s%s", s1, s2) != EOF)    {        int len = strlen(s1); CLR(dp, INF);        for(int i = len-1; i >= 0; i--)        {            for(int j = i; j < len; j++)            {                if(i == j)                {                    dp[i][j] = 1;                    continue;                }                dp[i][j] = dp[i+1][j]+1;                for(int k = i+1; k <= j; k++)                {                    if(s2[i] == s2[k])                    {                        if(k == j)                            dp[i][j] = min(dp[i][j], dp[i+1][j]);                        else                            dp[i][j] = min(dp[i][j], dp[i+1][k]+dp[k+1][j]);                    }                }            }        }        for(int i = 0; i < len; i++)            ans[i] = dp[0][i];        for(int i = 0; i < len; i++)        {            if(s1[i] == s2[i])            {                if(i == 0)                    ans[i] = 0;                else                    ans[i] = ans[i-1];            }            else            {                for(int j = 0; j < i; j++)                    ans[i] = min(ans[i], ans[j] + dp[j+1][i]);            }        }        Pi(ans[len-1]);    }    return 0;}