제출 #973920

#제출 시각아이디문제언어결과실행 시간메모리
973920GrindMachineFish (IOI08_fish)C++17
56 / 100
289 ms65536 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long int ll; typedef long double ld; typedef pair<int,int> pii; typedef pair<ll,ll> pll; #define fastio ios_base::sync_with_stdio(false); cin.tie(NULL) #define pb push_back #define endl '\n' #define sz(a) (int)a.size() #define setbits(x) __builtin_popcountll(x) #define ff first #define ss second #define conts continue #define ceil2(x,y) ((x+y-1)/(y)) #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define yes cout << "Yes" << endl #define no cout << "No" << endl #define rep(i,n) for(int i = 0; i < n; ++i) #define rep1(i,n) for(int i = 1; i <= n; ++i) #define rev(i,s,e) for(int i = s; i >= e; --i) #define trav(i,a) for(auto &i : a) template<typename T> void amin(T &a, T b) { a = min(a,b); } template<typename T> void amax(T &a, T b) { a = max(a,b); } #ifdef LOCAL #include "debug.h" #else #define debug(x) 42 #endif /* knew beforehand that this problem could be solved with segtree */ int MOD = 1e9 + 7; const int N = 1e5 + 5; const int inf1 = int(1e9) + 5; const ll inf2 = ll(1e18) + 5; void solve(int test_case) { ll n,m; cin >> n >> m >> MOD; vector<pll> a(n+5); rep1(i,n) cin >> a[i].ff >> a[i].ss; sort(a.begin()+1,a.begin()+n+1); vector<ll> cnt(m+5); rep1(i,n) cnt[a[i].ss]++; vector<ll> mnl(m+5,inf2), mxl(m+5,-inf2); rep1(i,n){ amin(mnl[a[i].ss],a[i].ff); amax(mxl[a[i].ss],a[i].ff); } vector<bool> came(m+5); ll can_see[m+5][m+5]; memset(can_see,-1,sizeof can_see); ll ptr = n; ll ans = 0; rev(i,n,1){ auto [len,x] = a[i]; while(ptr >= 1 and a[ptr].ff > len/2){ cnt[a[ptr].ss]--; ptr--; } if(came[x]) conts; came[x] = 1; // not maxed out ll prod = 1; rep1(y,m){ if(y != x and can_see[y][x] >= cnt[x]) conts; if(y != x){ prod = prod*(cnt[y]+1)%MOD; } else{ prod = prod*cnt[y]%MOD; } } ans = (ans+prod)%MOD; // maxed out prod = 1; rep1(y,m){ if(can_see[y][x] > cnt[x] or x == y) conts; prod = prod*(cnt[y]+1)%MOD; } ans = (ans+prod)%MOD; // upd can_see rep1(y,m){ can_see[x][y] = cnt[y]; } } cout << ans << endl; } int main() { fastio; int t = 1; // cin >> t; rep1(i, t) { solve(i); } return 0; }
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