uva 12545 Bits Equalizer

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”•
SisnowequaltoT

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

//
//  main.cpp
//  uva 12545 - Bits Equalizer
//
//  Created by XD on 15/8/16.
//  Copyright (c) 2015年 XD. All rights reserved.
//

/*
贪心法
首先注意到题中所给的变化，只能增加 1 的个数，所以onel1 与onel2的个数是必须相等的
因此得到了算法的主体
1 onel1 > onel2     输出-1，因为无论如何变化，第一个串中的1的个数始不会减少
2 onel1 == onel2    问号全部变成0,然后统计s1和s2不同处有多少个（设为num个），那么ans == 问号的个数 + num/2 。这里的num一定是偶数。很简单，串只是01串，那么某一处不同，则必存在另一处不同。
3. onel1 < onel2    先变化问号的地方，因为这样肯定不会增加最终的操作次数的。另外处变化成第二个字符串相应的字符（当然如果1的个数已经够了，则后面的问号全部变化为0）；接下来如果1的个数还没够，则将串一中的0变化成1，当然也是贪心法，也就是在串一位0串二为1的地方串一0-》1.

*/
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<vector>
#include <string.h>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <cstdio>
using namespace std ;
char s[110];
char s1[110] ;
int get_ch_num(char s[],int l,char ch )
{
int num =0  ;
for (int i = 0; i < l ; i++) {
if (s[i]==ch) {
num++ ;
}
}
return num ;
}
int main(int argc, const char * argv[]) {
int  T ;scanf("%d" ,&T) ;
int kase = 0;int ans;
while (T--) {
ans = 0  ;
scanf(" %s" ,s)  ;
scanf(" %s" ,s1) ;
printf("Case %d: ",++kase) ;
int len1 = (int)strlen(s) ;
int len2 = (int ) strlen(s1)  ;
int onel1 = get_ch_num(s,len1,'1') ;
int onel2 = get_ch_num(s1 , len2,'1') ;
if (onel1 > onel2) {
printf("-1\n")  ;
}
else if (onel1 == onel2)
{
ans += get_ch_num(s , len1,'?') ;
int  t =0 ;
for (int i = 0; i < len1; i++) {
if (s[i]=='?'   ) {
s[i]='0' ;
}
if (s[i]!=s1[i]) {
t++ ;
}
}
ans+=(t/2) ;
printf("%d\n",ans) ;
}
else{
int d = onel2 - onel1 ;

for (int i = 0 ; i < len1; i++) {
if (s[i] =='?') {
if (s1[i] == '1'&&d > 0 ) {
s[i] = '1' ;ans++ ;
d-- ;
}
else
{
ans++ ;
s[i] = '0' ;
}
}
}
if (d>0) {
for (int i = 0; i < len1; i++) {
if (s[i] == '0' && s1[i] == '1'&&d>0) {
s[i] = '1' ;d-- ;
ans++ ;
}
}
}
int  t =0 ;
for (int i = 0; i < len1; i++) {
if (s[i]!=s1[i]) {
t++ ;
}
}
ans+=(t/2) ;
printf("%d\n" ,ans) ;
}
}

return 0;
}