HDU-2476 String painter

 

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

题意：每次将第一个字符串一段连续区间赋予一个值，问从串一变到串二最少多少步。

思路：先求出每个位置都至少被染一次的方案数。dp[i][j]最坏为dp[i][j - 1] + 1；若在i到j - 1之间有k使s[k] == s[j]，则dp[i][j] = min(dp[i][j], dp[i][k - 1] + dp[k][j])，此时意义便是先染k到j，然后处理k + 1到j - 1。

求完dp后再求真正的答案。用ans[i]表示1到i之间需要染多少次，若s[i] == t[i], ans[i] = ans[i - 1]；否则枚举1到i - 1之间的k，ans[i] = min(ans[i], ans[k - 1] + dp[k][j])。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

char s[105], t[105];
int dp[105][105], ans[105];
int main() {
    while (~scanf("%s%s", s + 1, t + 1)) {
        int L = strlen(s + 1);
        clr(dp, 0);
        for (int l = 0; l < L; ++l) {
            for (int i = 1; i <= L - l; ++i) {
                int j = i + l;
                dp[i][j] = dp[i][j - 1] + 1;
                for (int k = i; k < j; ++k) {
                    if (t[k] == t[j]) dp[i][j] = min(dp[i][j], dp[i][k - 1] + dp[k][j - 1]);
                }
            }
        }
        /*for (int i = 1; i <= L; ++i) {
            for (int j = 1; j <= L; ++j) {
                cout << dp[i][j] << ' ';
            }
            cout << endl;
        }*/
        clr(ans, 63); ans[0] = 0;
        for (int i = 1; i <= L; ++i) {
            if (s[i] == t[i]) ans[i] = ans[i - 1];
            else {
                for (int k = 0; k < i; ++k) ans[i] = min(ans[i], ans[k] + dp[k + 1][i]);
            }
        }
        printf("%d
", ans[L]);
    }
    return 0;
}