UVA12545-Bits Equalizer（思维）

Problem UVA12545-Bits Equalizer

Accept: 821  Submit: 4548
Time Limit: 3000 mSec

Problem Description

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

00000

 

 Sample Output

Case 1: 3

Case 2: 1

Case 3: -1

题解：这个题还是挺不错的，结论很简洁，但是不是太好想。首先明确一点就是要进行几步操作和字符串的顺序无关，只和对应位置的对应情况有关，一共由四种对应：

1、0 --> 1

2、1 --> 0

3、？--> 1

4、？--> 0

分别记录个数为cnt[1~4]，首先cnt[3、4]是肯定要加进去的，只需考虑别的步骤，2这种对应只能通过交换来消除，所以如果cnt1+cnt3<cnt2，就是无解的，否则答案就是在原来的基础上加上max(cnt1, cnt2).

解释一下，如果cnt1 >= cnt2，那么只用1这种对应来交换的就已经足够了，还需要把多出来的1对应变成正确对应因此此时加上cnt1，如果cnt1 < cnt2，这时需要用3对应来配，此时需要cnt2步操作，因此就是在cnt[3、4]的基础上加上二者最大值。

#include <bits/stdc++.h>

using namespace std;

int T = 1;

int main()
{
    //freopen("input.txt", "r", stdin);
    int iCase;
    scanf("%d", &iCase);
    while (iCase--) {
        string ori, tar;
        cin >> ori >> tar;

        int cnt = 0, cnt2 = 0, cnt3 = 0, cnt4 = 0;
        int len = ori.length(); 
        for (int i = 0; i < len; i++) {
            if (ori[i] == '0' && tar[i] == '1') {
                cnt++;
            }
            else if (ori[i] == '1' && tar[i] == '0') {
                cnt2++;
            }
            else if (ori[i] == '?') {
                if (tar[i] == '1') cnt3++;
                else cnt4++;
            }
        }

        if (cnt + cnt3 < cnt2) {
            printf("Case %d: %d
", T++, -1);
        }
        else {
            int ans = cnt3 + cnt4;
            ans += max(cnt, cnt2);
            printf("Case %d: %d
", T++, ans);
        }
    }
    return 0;
}