uva12545 Bits Equalizer

uva12545 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 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

 

这个题目用到的是统计的思想，不用真正的去变字符串里字符的位置，只需要统计哪几个要变动就OK。

 

#include <algorithm>

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

const int MAXN = 255;

int num[4];

char S[MAXN], T[MAXN];

int solve(int len)

{

    memset(num, 0, sizeof(num));

    for(int i = 0; i < len; ++i)

    {

        if(S[i] == '0' && T[i] == '1' ) num[0]++;

        else if(S[i] == '1' && T[i] == '0') num[1]++;

        else if(S[i] == '?' && T[i] == '1') num[2]++;

        else if(S[i] == '?' && T[i] == '0') num[3]++;

    }



    int ans = min(num[0], num[1]);

    if(num[0] >= num[1])

    {

        num[0] -= ans;

        if(num[3] - num[0] >= 0)

        {

            ans += 2*num[0];

            ans += num[2]  + num[3] - num[0];

        }

        else

        {

            ans += 2*num[3];

            ans += num[2] + num[0] - num[3];

        }

    }

    else if(num[0] < num[1])

    {

        num[1] -= ans;

        if(num[2] - num[1] >= 0)

        {

            ans += 2*num[1];

            ans += num[3] + num[2] - num[1];

        }

    }

    return ans;

}

int main()

{

    int Tcase, ans;

    int tag1, tag2;



    scanf("%d%*c", &Tcase);

    for(int t = 1; t <= Tcase; ++t)

    {

        tag1 = 0, tag2 = 0;

        ans = 0;

        gets(S);

        gets(T);

        int len = strlen(S);



        for(int i = 0; i < len; ++i)

        {

            if(S[i] == '1')   tag1++;

            if(T[i] == '1')   tag2++;

        }

        if(tag1 > tag2 )

            printf("Case %d: %d\n", t, -1);

        else

        {

            printf("Case %d: %d\n", t, solve(len));

        }

    }

    return 0;

}