Submission #907542

#	Time	Username	Problem	Language	Result	Execution time	Memory
907542		sysia	Star Trek (CEOI20_startrek)	C++17	0 / 100	29 ms	652 KiB

startrek

//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;

void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

#define int long long
typedef pair<int, int> T;
const int oo = 1e18, oo2 = 1e9+7, K = 30;
const int mod = 998244353;

void solve(){
    int n, d; cin >> n >> d;
    //n <= 100, d = 1;
    vector<vector<int>>g(n+1);
    for (int i = 1; i<n; i++){
        int a, b; cin >> a >> b;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    vector dp(n+1, vector<int>(2));
    //jak jestesmy w lisciu w tak ukorzenionym drzewie to przegrywamy
    vector<int>depth(n+1), par(n+1);
    function<void(int, int)>dfs = [&](int v, int pa){
        par[v] = pa;
        dp[v][0] = 0;
        dp[v][1] = 1; //gracz 1 jest teraz w v i musi zrobic ruch, ale nie ma gdzie :p
        for (auto x: g[v]){
            if (x == pa) continue;
            depth[x] = depth[v]+1;
            dfs(x, v);
            dp[v][0] |= dp[x][1]; //musi byc jakas wygrywajaca
            dp[v][1] &= (dp[x][0] == 1); //musi prowadzic do samych wygrywajacych
        }
        debug(v, dp[v]);
    };
    vector win(n+1, vector<int>(2));
    for (int i = 1; i<=n; i++){
        dp.assign(n+1, vector<int>(2, 0));
        depth.assign(n+1, 0);
        dfs(i, i);
        win[i] = dp[i];
        // return;
    }
    depth[1] = 0;
    dfs(1, 1);
    int ans = 0;
    for (int i = 1; i<=n; i++){
        for (int j = 1; j<=n; j++){
            //krawedz i--j
            if (depth[i]%2 == 0){
                if (win[1][0] || win[j][1]) {
                    ans++;
                }
            } else {
                //gracz 2 moze podejmowac decyzje
                //nie dojdziemy w jakis sposob do wierzcholka i(?)
                //zeby dojsc do wierzcholka i, na calej sciezce od i do 1 gracz 2 musi wykonac decyzje, ze warto isc w te konkretne poddrzewo
                int v = i;
                int prev = -1;
                bool ok = 1;
                while (v != 1){
                    for (auto x: g[v]){
                        if (x == prev || x == par[v]) continue;
                        debug(v, x);
                        if (!dp[x][0]) {
                            //zmieniam zdanie, ide w to poddrzewo
                            ok = 0;
                            break;
                        }
                    }
                    v = par[v];
                    prev = v;
                    v = par[v];
                }
                if (ok && win[j][0]){
                    debug(i, j);
                    ans++;
                }
            }
        }
    }
    cout << ans << "\n";
}

int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    //cin >> t;
    while (t--) solve();

    return 0;
}

• Subtask 1: 0 / 0
• Subtask 2: 0 / 7
• Subtask 3: 0 / 8
• Subtask 4: 0 / 15
• Subtask 5: 0 / 15
• Subtask 6: 0 / 20
• Subtask 7: 0 / 20
• Subtask 8: 0 / 15