UVA12545 Bits Equalizer(脑洞)

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 ofmoves. 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 caseconsists of two lines. The first line is the string S consisting of ‘0’, ‘1’ and ‘?’. The second line is
thestring 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 toconvert 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

给出两个字符串，问你最小变换次数使得两字符串相同。合理的变换有：0变为1，?变为0或1，交换两个相邻字符。

统计题目给出的符合合理变换的数目，包括num1计数"?1"，num0计数"?0" ，zero计数 "01" ，one计数"10"。

如果"01"数目加上"?1"数目少于"10"数目，则不能通过交换使两个字符串相同。

如果"01"数目大于等于"10"数目则变换次数为zero + num0 + num1。

否则变换次数为one + num0 + num1。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 105;
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d", &t);
	for(int cas = 1; cas <= t; ++cas) {
		char s1[MAXN], s2[MAXN];
		scanf("%s%s", s1, s2);
		int len1 = strlen(s1), len2 = strlen(s2);
		if(len1 != len2) {
			printf("Case %d: -1\n", cas); 
			continue;
		}	
		int zero = 0, one = 0, num1 = 0, num0 = 0;
		for(int i = 0; i < len1; ++i) {
			if(s1[i] == '?') {
				if(s2[i] == '1') num1++;
				if(s2[i] == '0') num0++;
			}
			if(s1[i] == '0' && s2[i] == '1') zero++;
			if(s1[i] == '1' && s2[i] == '0') one++;
		}
		if(zero >= one) printf("Case %d: %d\n", cas, zero + num0 + num1);
		else if(zero + num1 < one) printf("Case %d: -1\n", cas); // 无法通过交换满足题意
		else printf("Case %d: %d\n", cas, one + num1 + num0);
	}
	return 0;
}