zoukankan      html  css  js  c++  java
  • Codeforces Gym 100203E E

    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
    ",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
    ",cas,ans+tmp/2);
        }
    }
  • 相关阅读:
    面向对象编程,其属性特性,用法等
    re正则模块细解,面向对象编程思路的优劣。
    机器人学——1.2-三维空间位姿描述
    机器人学——1.1-二维空间位姿描述
    机器人学——1.0-位置与姿态概述
    latex教程:1.2-latex现状
    latex教程: 1.1-历史
    windows安装opencv
    使用pip安装Opencv
    在Ubuntu上安装opencv-python
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4733289.html
Copyright © 2011-2022 走看看