【题目链接】:http://codeforces.com/contest/779/problem/E

【题意】



给你n个长度为m的二进制数

(有一些是通过位运算操作两个数的形式给出);

然后有一个未知数字,用符号’?’表示;

(所有的数字都是长度为m的二进制数);

然后让你确定这个’?’使得n个数字的和最大、最小;

【题解】



模拟题;

会有嵌套的情况；

即

a=x & y

b=a&?

类似这样.

所以在算b之前,先把a的值确定下来;

明白了怎么算之后;

枚举那个未知数字的m位的每一位;

每一位都只有两种可能,即为0或者为1;

假设为0;

然后带进去算一下最后结果,这一位的和为多少;

假设为1

然后带进去算一下最后结果,这一位的和为多少;

如果要最大值就选那个大的对应的数字,最小值则相反。



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define ps push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%lld",&x)

#define ref(x) scanf("%lf",&x)

typedef pair<int, int> pii;

typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };

const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };

const double pi = acos(-1.0);

const int M = 1e3+100;

const int N = 5e3 + 100;

struct abc

{

    int a, b, p;

    int sz[M];

};

int n, m,change,f[N];

string name,s;

map <string, int> dic;

abc a[N];

vector<int>v1, v2;

int get_val(int pos)

{

    int sum = 0;

    f[0] = change;

    rep1(i, 1, n)

    {

        if (a[i].p == 0)

        {

            sum += a[i].sz[pos];

            f[i] = a[i].sz[pos];

            continue;

        }

        int x = f[a[i].a], y = f[a[i].b];

        int p = a[i].p;

        if (p == 1)

            f[i] = x&y;

        if (p == 2)

            f[i] = x | y;

        if (p == 3)

            f[i] = x^y;

        sum += f[i];

    }

    return sum;

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    rei(n), rei(m);

    dic["?"] = 0;

    rep1(i, 1, n)

    {

        cin >> name; dic[name] = i;

        cin >> s; cin >> s;

        if (s[0] == '1' || s[0] == '0')

        {

            a[i].p = 0;

            rep1(j, 1, m)

                a[i].sz[j] = s[j - 1]-'0';

        }

        else

        {

            a[i].a = dic[s];

            cin >> s;

            if (s[0] == 'A') a[i].p = 1;

            if (s[0] == 'O') a[i].p = 2;

            if (s[0] == 'X') a[i].p = 3;

            cin >> s;

            a[i].b = dic[s];

        }

    }

    rep1(i, 1, m)

    {

        change = 0; int temp0 = get_val(i);

        change = 1; int temp1 = get_val(i);

        if (temp0 <= temp1)

            v1.ps(0);

        else

            v1.ps(1);

        if (temp0 >= temp1)

            v2.ps(0);

        else

            v2.ps(1);

    }

    rep1(i, 0, m - 1)

        printf("%d", v1[i]);

    puts("");

    rep1(i, 0, m - 1)

        printf("%d", v2[i]);

    puts("");

    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);

    return 0;

}