首页 > ACM题库 > HDU-杭电 > hdu 2476 String painter-动态规划-[解题报告]C++
2014
01-26

hdu 2476 String painter-动态规划-[解题报告]C++

String painter

问题描述 :

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?

输入:

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.

输出:

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.

样例输入:

zzzzzfzzzzz
abcdefedcba
abababababab
cdcdcdcdcdcd

样例输出:

6
7

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int MAXN=110;
int dp[MAXN][MAXN];
char str1[MAXN],str2[MAXN];
int ans[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%s%s",str1,str2)==2)
    {
        int n=strlen(str1);
        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;
        //先直接DP求出从空白串变成str2
        for(int i=n-2;i>=0;i--)
            for(int j=i+1;j<n;j++)
            {
                dp[i][j]=dp[i+1][j]+1;
                for(int k=i+1;k<=j;k++)
                    if(str2[i]==str2[k])
                        dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);
            }
        for(int i=0;i<n;i++)
        {
            ans[i]=dp[0][i];
            if(str1[i]==str2[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;
}

解题转自:http://www.cnblogs.com/kuangbin/archive/2013/04/30/3052043.html


  1. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  2. 题目需要求解的是最小值,而且没有考虑可能存在环,比如
    0 0 0 0 0
    1 1 1 1 0
    1 0 0 0 0
    1 0 1 0 1
    1 0 0 0 0
    会陷入死循环

  3. 如果两个序列的最后字符不匹配(即X [M-1]!= Y [N-1])
    L(X [0 .. M-1],Y [0 .. N-1])= MAX(L(X [0 .. M-2],Y [0 .. N-1]),L(X [0 .. M-1],Y [0 .. N-1])
    这里写错了吧。