Submission #211939

#	Time	Username	Problem	Language	Result	Execution time	Memory
211939		tatyam	Flood (IOI07_flood)	C++17	81 / 100	2091 ms	12780 KiB

flood

#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using tuplis = array<int, 3>;
#define rep(b) for(int i = 0; i < b; i++)
#define each(i,a) for(auto&& i : a)
#define all(a) begin(a), end(a)
 
 
struct wall{
    int to = -1, index;
    wall(){}
    wall(int to, int index): to(to), index(index){}
    bool exist() const { return to + 1; }
};
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<pii> p(n);
    each(i, p) cin >> i.first >> i.second;
    set<tuplis> min_p;
    rep(n) min_p.insert({p[i].first, p[i].second, i});
    vector<array<wall, 4>> g(n);
    auto submerge = [&](int i) -> void {
        if(!any_of(all(g[i]), [](const wall& w){ return w.exist(); })) min_p.erase({p[i].first, p[i].second, i});
    };
    int w;
    cin >> w;
    rep(w){
        int a, b;
        cin >> a >> b;
        a--; b--;
        int d = 0;
        if(p[a].first < p[b].first) d = 0;
        else if(p[a].first > p[b].first) d = 2;
        else if(p[a].second < p[b].second) d = 1;
        else d = 3;
        g[a][d] = {b, i};
        g[b][d ^ 2] = {a, i};
    }
    vector<bool> ans(w, 1);
    while(min_p.size()){
        int start = (*min_p.begin())[2], at = start, d = 0;
        vector<pii> visit;
        if(!g[at][d].exist()) d = 1;
        do{
            if(!g[at][d].exist()){
                d++;
                d &= 3;
                continue;
            }
            visit.emplace_back(at, d);
            ans[g[at][d].index] = !ans[g[at][d].index];
            at = g[at][d].to;
            d--;
            d &= 3;
        }while(at != start);
        each(i, visit){
            int at, d;
            tie(at, d) = i;
            if(!g[at][d].exist()) continue;
            int at2 = g[at][d].to, d2 = d ^ 2;
            g[at][d].to = -1;
            g[at2][d2].to = -1;
            submerge(at);
            submerge(at2);
        }
    }
    cout << accumulate(ans.begin(), ans.end(), 0) << '\n';
    rep(w) if(ans[i]) cout << i + 1 << '\n';
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 3 / 3
• Subtask 5: 0 / 10
• Subtask 6: 3 / 3
• Subtask 7: 3 / 3
• Subtask 8: 3 / 3
• Subtask 9: 3 / 3
• Subtask 10: 0 / 9
• Subtask 11: 9 / 9
• Subtask 12: 9 / 9
• Subtask 13: 9 / 9
• Subtask 14: 9 / 9