UVA 12545 Bits Equalizer

题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=27865

Bits Equalizer

Time Limit: 1000ms

Memory Limit: 131072KB

 

This problem will be judged on UVA. Original ID: 12545
64-bit integer IO format: %lld      Java class name: Main

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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 convertS 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 (C200) 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

Source

Western and Southwestern European Regionals, 2012 - Valencia

 

题目大意：将第一个字符与第二个字符每一个位置都相对应，至少需要移动的次数为多少。有以下的规则：1、0可以变成1；2、？可以变成0 or 1；3、任意0与1的位置可以互换

 

解题思路：统一0和1以及？的个数，因为？一定要变换，所以最少移动次数也得有？个数次，所以在其基础上加就可以了。

分两种情况考虑：（交换、变换）

优先变换情况：

1、存在1/0，并且后面可以找到0/1，那么交换。

2、存在1/0，但是找不到与其对应的0/1可以交换，那么就只能选择？/1,那么交换。

变换情况：

1、存在0/1，这种情况的话就直接把0变换成1即可。

 

综上，只要把所有情况考虑到，AC就不再是梦想啦，哇哈哈~~

 

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main ()
{
    int n,len1,flag=1;
    char s[110],t[110];
    scanf("%d",&n);
    while (n--)
    {
        scanf("%s%s",s,t);
        len1=strlen(s);
        //len2=strlen(t);
        int a,b,c,aa,bb;
        a=b=c=aa=bb=0;
        for (int i=0; i<len1; i++)
        {
            if (s[i]=='0')
                a++;
            else if (s[i]=='1')
                b++;
            else
                c++;
        }
        for (int i=0; i<len1; i++)
        {
            if (t[i]=='0')
                aa++;
            else
                bb++;
        }
        if (b>bb)
        {
            printf ("Case %d: -1
",flag++);
            continue;
        }
        int sum=0;
        int j;
        if (s[i]=='1'&&t[i]=='0')
        {
            for (j=0; j<len1; j++)
            {
                if (s[j]=='0'&&t[j]=='1')
                {
                    sum++;
                    s[i]='0';
                    s[j]='1';
                    break;
                }
            }
            if (j>=len1)
                break;

        }
    }
    for (int i=0; i<len1; i++)
    {
        int j;
        if (s[i]=='1'&&t[i]=='0')
        {
            for (j=0; j<len1; j++)
            {
                if (s[j]=='?'&&t[j]=='1')
                {
                    sum++;
                    s[i]='?';
                    s[j]='1';
                    break;
                }
            }
            if (j>=len1)
                break;
        }
    }
    for (int i=0; i<len1; i++)
    {
        if (s[i]=='0'&&t[i]=='1')
            sum++;
        //ans=sum+c;
    }
    printf ("Case %d: %d
",flag++,sum+c);
}
return 0;
}