Codeforces Gym 100203E E - bits-Equalizer 贪心

最新推荐文章于 2019-09-03 00:38:44 发布

转载最新推荐文章于 2019-09-03 00:38:44 发布 · 150 阅读

文章标签：

#数据结构与算法

本文探讨了在给定字符串转换任务中，如何利用贪心算法优化代码效率，通过处理问号、零到一的转换及字符交换来实现字符串从S到T的最小步骤转换。重点介绍了每一步的操作策略及其背后的逻辑。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

E - bits-Equalizer
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87794#problem/K

Description

You are given two non-empty strings S and T of equal lengths. S contains the characters 0, 1 and ?, whereas T contains 0 and 1 only. Your task is to convert S into T in minimum number of moves. In each move, you can:

1. change a 0 in S to 1

2. change a ? in S to 0 or 1

3. swap any two characters in S

As an example, suppose S = 01??00 and T = 001010. We can transform S into T in 3 moves:

• Initially S = 01??00

• Move 1 – change S[2] to 1. S becomes 011?00

• Move 2 – change S[3] to 0. S becomes 011000

• Move 3 – swap S[1] with S[4]. S becomes 001010

• S is now equal to T

Input

The first line of input is an integer C (C ≤ 200) that indicates the number of test cases. Each case consists of two lines. The first line is the string S consisting of ‘0’, ‘1’ and ‘?’. The second line is the string T consisting of ‘0’ and ‘1’. The lengths of the strings won’t be larger than 100.

Output

For each case, output the case number first followed by the minimum number of moves required to convert S into T. If the transition is impossible, output - 1 instead.

Sample Input

3
01??00
001010
01
10
110001
000000

Sample Output

Case 1: 3
Case 2: 1
Case 3: -1

HINT

题意

给你一个s1和s2，你每次有三种操作，第一种是将0变成1，第二种是把问号变成1或者0，第三种是交换任意两个字符的位置

题解：

贪心，首先处理问号，然后再处理0变成1的问题，最后处理交换位置的问题

每次处理都直接扫一遍就好了

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

string s1,s2;
int num1x,num1y,num2x,num2y;
int main()
{
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        num1x=num1y=num2x=num2y=0;
        cin>>s2>>s1;
        int len=s1.size();
        for(int i=0;i<len;i++)
        {
            if(s1[i]=='1')
                num1x++;
            if(s1[i]=='0')
                num1y++;
            if(s2[i]=='1')
                num2x++;
            if(s2[i]=='0')
                num2y++;
        }
        int flag=0;
        int ans=0;
        for(int i=0;i<len;i++)
        {
            if(num2x>num1x)
            {
                flag=1;
                break;
            }
            if(s2[i]=='?')
            {
                if(s1[i]=='1')
                {
                    if(num2x<num1x)
                    {
                        s2[i]='1';
                        num2x++;
                        ans++;
                    }
                    else
                    {
                        s2[i]='0';
                        num2y++;
                        ans++;
                    }
                }
                else
                {
                    if(num2y<num1y)
                    {
                        s2[i]='0';
                        num2y++;
                        ans++;
                    }
                    else
                    {
                        s2[i]='1';
                        num2x++;
                        ans++;
                    }
                }
            }
        }
        if(flag==1)
        {
            printf("Case %d: -1\n",cas);
            continue;
        }
        for(int i=0;i<len;i++)
        {
            if(num1x==num2x&&num2x==num2y)
                break;
            if(s2[i]=='0'&&s1[i]=='1')
            {
                if(num1x>num2x)
                {
                    num2y--;
                    num2x++;
                    s2[i]='1';
                    ans++;
                }
            }
        }
        int tmp=0;
        for(int i=0;i<len;i++)
        {
            if(s1[i]!=s2[i])
            {
                tmp++;
            }
        }
        printf("Case %d: %d\n",cas,ans+tmp/2);
    }
}