제출 #388161

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
388161		Sara	Plus Minus (BOI17_plusminus)	C++14	100 / 100	139 ms	18360 KiB

plusminus

#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;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define ll long long
#define F first
#define S second
#define pb push_back
#define lc (ind << 1)
#define rc (lc | 1)
#define md ((b + e) >> 1)

const ll N = 1e5 + 5;
const ll M = 1e6 + 5;
const ll LOG = 30;
const ll INF = 1e9 + 5;
const ll MOD = 1e9 + 7;

ll power(ll x, ll t){
    if (t == 0){
        return 1ll;
    }
    return ((t & 1) ? x : 1ll) * power(x * x % MOD, t >> 1ll) % MOD;
}

int n, m, k;
int X[N];
int Y[N];
bool ty[N];
vector <int> bib1, bib2;
int ind1[N], ind2[N];
vector <int> in1[2][N], in2[2][N];

inline void compress(vector <int> &vec){
    sort(vec.begin(), vec.end());
    vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
    return;
}

inline ll solve(){
    if (k == 0){
        return (2ll * power(2ll, m - 1ll)  % MOD + (power(2ll, n) - 2ll)) % MOD;
    }
    ll bad = 1;
    ll ans1 = power(2ll, (ll)n - (ll)bib1.size());
    for (int i = 0; i < (int)bib1.size(); i ++){
        for (int j = 1; j < (int)in1[0][i].size(); j ++){
            bool f1 = (in1[0][i][j - 1] & 1);
            bool f2 = (in1[0][i][j] & 1);
            if (f1 != f2){
                ans1 = 0;
                bad = 0;
            }
        }
        for (int j = 1; j < (int)in1[1][i].size(); j ++){
            bool f1 = (in1[1][i][j - 1] & 1);
            bool f2 = (in1[1][i][j] & 1);
            if (f1 != f2){
                ans1 = 0;
                bad = 0;
            }
        }
        if ((int)in1[0][i].size() && (int)in1[1][i].size()){
            bool f1 = (in1[0][i][0] & 1);
            bool f2 = (in1[1][i][0] & 1);
            if (f1 == f2){
                ans1 = 0;
                bad = 0;
            }
        }
    }
    ll ans2 = power(2ll, (ll)m - (ll)bib2.size());
    for (int i = 0; i < (int)bib2.size(); i ++){
        for (int j = 1; j < (int)in2[0][i].size(); j ++){
            bool f1 = (in2[0][i][j - 1] & 1);
            bool f2 = (in2[0][i][j] & 1);
            if (f1 != f2){
                ans2 = 0;
                bad = 0;
            }
        }
        for (int j = 1; j < (int)in2[1][i].size(); j ++){
            bool f1 = (in2[1][i][j - 1] & 1);
            bool f2 = (in2[1][i][j] & 1);
            if (f1 != f2){
                ans2 = 0;
                bad = 0;
            }
        }
        if ((int)in2[0][i].size() && (int)in2[1][i].size()){
            bool f1 = (in2[0][i][0] & 1);
            bool f2 = (in2[1][i][0] & 1);
            if (f1 == f2){
                ans2 = 0;
                bad = 0;
            }
        }
    }
    return ((ans1 + ans2) % MOD - bad + MOD) % MOD;
}


int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    cin >> n >> m >> k;
    for (int i = 0; i < k; i ++){
        char c;
        cin >> c >> X[i] >> Y[i];
        ty[i] = (c == '-');
        bib1.pb(X[i]); bib2.pb(Y[i]);
    }
    compress(bib1); compress(bib2);
    for (int i = 0; i < k; i ++){
        ind1[i] = lower_bound(bib1.begin(), bib1.end(), X[i]) - bib1.begin();
        ind2[i] = lower_bound(bib2.begin(), bib2.end(), Y[i]) - bib2.begin();
    }
    for (int i = 0; i < k; i ++){
        in1[ty[i]][ind1[i]].pb(Y[i]);
        in2[ty[i]][ind2[i]].pb(X[i]);
    }
    cout << (solve() % MOD + MOD) % MOD << '\n';
    return 0;
}
/*
2 4 4
+ 1 1
- 1 2
+ 1 3
- 1 4
*/
/*
3 3 3 
- 2 1
+ 2 3
+ 3 3
*/

• Subtask 1: 12 / 12
• Subtask 2: 42 / 42
• Subtask 3: 46 / 46