Submission #1108126

#TimeUsernameProblemLanguageResultExecution timeMemory
1108126laightTropical Garden (IOI11_garden)C++17
0 / 100
4 ms14160 KiB
#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], vis2[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){ queue<pi> q; q.push({0, c}); while (!q.empty()){ int cur = q.front().se, cd = q.front().fi; if (vis[cur]) continue; vis[cur] = 1; q.pop(); if (cur < n) dis.pb(cd); for (int i : rev[c]){ if (!vis[i]){ q.push({cd+1, i}); } } } } void g_f2(int c){ queue<pi> q; q.push({0, c}); while (!q.empty()){ int cur = q.front().se, cd = q.front().fi; if (vis2[cur]) continue; vis2[cur] = 1; q.pop(); if (cur < n) dis2.pb(cd); for (int i : rev[c]){ if (!vis2[i]){ 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); } 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); g_f2(n+p); for (int i=0; i<q; ++i){ int tv=g[i], ca=0; for (int j : dis){ if (j > tv) break; if (cy1){ if (tv >= j && tv%cl1 == j%cl1) ++ca; } else{ if (tv == j) ++ca; } } for (int j : dis2){ if (j > tv) break; if (cy2){ if (tv >= j && tv%cl2 == j%cl2) ++ca; } else{ if (tv == j) ++ca; } } answer(ca); } }
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