제출 #57894

#제출 시각아이디문제언어결과실행 시간메모리
57894BenqIli (COI17_ili)C++14
49 / 100
4019 ms44472 KiB

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
 
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
 
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
 
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 10001;
 
int n,m, status[MX];
bitset<MX> res[MX], posi, bad[MX];
vi z;
vi toCheck[MX];
 
bitset<MX> clas(string a, int t) {
    int x = stoi(a.substr(1,sz(a)-1));
    if (a[0] == 'x') {
        bitset<MX> z;
        z[x] = 1;
        return z;
    }
	if (bad[x] == 1) bad[t] = 1;
    return res[x];
}
 
bool check(int x) {
	for (int i: z) {
		bool ok = 1;
		for (int b: toCheck[i]) {
			if (!res[x][b]) {
				ok = 0;
				break;
			}
		}
		if (ok) return 1;
	}
	return 0;
}
 
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> m;
    FOR(i,1,m+1) {
        char c; cin >> c;
        if ('0' <= c && c <= '1') status[i] = c-'0';
        else status[i] = -1;
    }
    FOR(i,1,m+1) {
        string a,b; cin >> a >> b;
        res[i] = clas(a,i)|clas(b,i);
        if (status[i] == 0) posi |= res[i];
        if (status[i] == 1) {
        	if (bad[i] == 0) z.pb(i);
        	bad[i] = 1;
        }
    }
    posi.flip();
    FOR(i,1,m+1) res[i] &= posi;
    for (int i: z) {
    	FOR(j,1,n+1) if (res[i][j]) toCheck[i].pb(j);
    }
    FOR(i,1,m+1) if (status[i] == -1) {
    	if (res[i].count() == 0) {
    		status[i] = 0;
    		continue;
    	}
    	if (check(i)) status[i] = 1;
    }
    FOR(i,1,m+1) {
    	if (status[i] == -1) cout << "?";
    	else cout << status[i];
    }
}
 
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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