【codeforces 779E】Bitwise Formula

【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;
}