Submission #1108131

#	Time	Username	Problem	Language	Result	Execution time	Memory
1108131		laight	Tropical Garden (IOI11_garden)	C++17	0 / 100	4 ms	14160 KiB

garden

#include "garden.h"
#include "gardenlib.h"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define pi pair<int, int>
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define ld long double
#define pb push_back
#define mp make_pair
#define pll pair<ll, ll>
#define umi unordered_map<int, int>
#define umll unordered_map<ll, ll>
#define umllv unordered_map<ll, vector<ll>>
 
#define getchar() (*_pinp?*_pinp++:(_inp[fread(_pinp=_inp, 1, 4096, stdin)]='\0', *_pinp++))
char _inp[4097], *_pinp=_inp;
#define scan(x) do{while((x=getchar())<'-'); _ssign=x=='-'; if(_ssign) while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0'); x=_ssign?-x:x;}while(0)
char _; bool _ssign;
 
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx2")
 
using namespace std; using namespace __gnu_pbds;
 
const int MM = 1e5+5e4+5;
const ll inf = 2e15+1;
const int lg = 17; // 17 for 1e5, 31 for 1e9
 
int n, m, p, a, b, lv, cycl, cl1, cl2, q, inp; bool cvis[2*MM], vis[2*MM], hcy, cy1, cy2;
vector<int> g(2005), ans(2005), to(2*MM), rev[2*MM], graph[MM], dis, dis2;
 
void g_f(int c, vector<int> &dv){
    queue<pi> q; q.push({0, c}); vis[c] = 1;
    while (!q.empty()){
        int cur = q.front().se, cd = q.front().fi;
        q.pop();
      	vis[cur] = 1;
        if (cur < n) dv.pb(cd);
        for (int i : rev[cur]){
            if (!vis[i]){
                vis[i] = 1;
                q.push({cd+1, i});
            }
        }
    }
}
void cyc_f(int c, int d, int pe){
    if (cvis[c]){
        if (c == pe) {hcy = 1; cycl = d;}
        return;
    }
    cvis[c] = 1; cyc_f(to[c], d+1, pe);
}
void count_routes(int N, int M, int P, int R[][2], int Q, int G[]){
    // cin.tie(0); cin.sync_with_stdio(0);
    // freopen("changyue_tst.txt", "r", stdin);
    n = N; m = M; p = P;
    for (int i=0; i<m; ++i){
        graph[R[i][0]].pb(R[i][1]); graph[R[i][1]].pb(R[i][0]);
    }
    q = Q; for (int i=0; i<q; i++) {g[i] = G[i];}
    for (int i=0; i<n; ++i){
        int oth = graph[i][0];
        if (i == graph[oth][0]){
            to[i] = n+oth;
        }
        else {to[i] = oth;}
        if (graph[i].size() >= 2){
            int oth2 = graph[i][1];
            if (i == graph[oth2][0]){
                to[n+i] = n+oth2;
            }
            else {to[n+i] = oth2;}
        }
        else{
            if (i == graph[oth][0]){
                to[n+i] = n+oth;
            }
            else {to[n+i] = oth;}
        }
    }
    for (int i=0; i<2*n; i++){
        rev[to[i]].pb(i);
    }
    cl1 = cl2 = 0;
    hcy = 0; cycl = 0;
    cyc_f(p, 0, p);
    cl1 = cycl; cy1 = hcy;
    memset(cvis, 0, sizeof(cvis));
    hcy = 0; cycl = 0;
    cyc_f(n+p, 0, n+p);
    cl2 = cycl; cy2 = hcy;
    g_f(p, dis); memset(vis, 0, sizeof(vis)); g_f(n+p, dis2);
    for (int i=0; i<q; ++i){
        int tv=g[i], ca=0;
        for (int j : dis){
            if (j > tv) break;
            if (cl1){
                if (!(tv-j)%cl1) ++ca;
            }
            else{
                if (tv == j) ++ca;
            }
        }
        for (int j : dis2){
            if (j > tv) break;
            if (cl2){
                if (!(tv-j)%cl2) ++ca;
            }
            else{
                if (tv == j) ++ca;
            }
        }
        answer(ca);
    }
}

• Subtask 1: 0 / 49
• Subtask 2: 0 / 20
• Subtask 3: 0 / 31