Codeforces779E-E. Bitwise Formula 【模拟】

对于 “？”的每一位，可以单独拆开来看。
所以对于”？”每一位枚举0,还是1，统计由？导出的其余变量，当前取0多还是取1多。取1个数多，最大。取0个数多，最小。
分析样例1
a := 101

b := 011

c := ? XOR b
预处理一下那些量跟？有关

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;
map<string,int> id;
map<string,int> opp;
struct node
{
    int id1,id2;
    int op;
    int in;
    int tt;
    string v;
}k[6000];
vector<int>to[6000];
vector<int>ids;
int ans1[2000];
int ans2[2000];
void topp(int i)
{
        if (i!=0) ids.push_back(i);
        for (int j = 0;j<to[i].size();j++)
        {
            k[to[i][j]].in--;
            if (k[to[i][j]].in==0) topp(to[i][j]);
        }
}
int main()
{
    int n,m;
    ios::sync_with_stdio(false);
    cin >> n >> m;
    string s1,ss,s2,so,s3;
    id["?"] = 0;
    k[0].in = 1;

    opp["AND"] = 1;
    opp["OR"] = 2;
    opp["XOP"] = 3;
    string sp;
    for (int i = 1; i<=m;i++)
    {
        sp+=" ";
    }
    k[0].v=sp;
    for (int i = 1; i<=n; i++)
    {
        cin >> s1 >> ss >> s2;
        id[s1] = i;
        k[i].in = 0;
        if (s2[0]!='0'&&s2[0]!='1')
        {
           cin >> so >> s3;
           k[i].tt = 0;
           k[i].v = sp;
           k[i].op = opp[so];
           k[i].id1 = id[s2];
           k[i].id2 = id[s3];
           if (k[id[s2]].in != 0)
           {
               k[i].in++;
               to[id[s2]].push_back(i);
           }
           if (k[id[s3]].in != 0)
           {
               k[i].in++;
               to[id[s3]].push_back(i);
           }
        }
        else
        {
            k[i].v = s2;
            k[i].tt = 1;
        }
    }
    for (int i = 1 ; i <= n ; i++ )
    {
        if (k[i].in==0)
            ids.push_back(i);
    }
    topp(0);
    int ans[2];
    int tans;
    for (int i = 0; i<m; i++)
    {
        for (int kk = 0; kk<=1; kk++)
        {
            k[0].v[i] = '0'+kk;
            ans[kk] = 0;
            for (int j = 0; j<ids.size(); j++)
            {
               if (k[ids[j]].tt==1)
               {
                   ans[kk]+=(k[ids[j]].v[i]-'0');
                   continue;
               }
               int v1 = k[k[ids[j]].id1].v[i]-'0';
               int v2 = k[k[ids[j]].id2].v[i]-'0';
               if (k[ids[j]].op==1)
               {
                    tans = v1 & v2;
               }
               else if (k[ids[j]].op==2)
               {
                   tans = v1|v2;
               }
               else
               {
                   tans = v1^v2;
               }
               ans[kk]+=tans;
               k[ids[j]].v[i] = tans+'0';
            }
        }
        if (ans[0]>ans[1])
        {
            ans1[i] = 1;ans2[i] = 0;
        }
        else if (ans[1]>ans[0])
        {
            ans1[i] = 0; ans2[i] = 1;
        }
        else ans1[i] = ans2[i] = 0;
    }
    for (int i = 0 ;i <m; i++)
        cout <<ans1[i];
    cout <<endl;
    for (int i = 0 ;i <m; i++)
        cout <<ans2[i];
    cout <<endl;
}