提出 #67002835

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ソースコード 拡げる

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
 
#define all(x) (x).begin(), (x).end()
 
template<template<typename> class Container, typename G>
ostream& operator<<(ostream& os, Container<G> x) {
    int f = 0; os << '{'; for (auto &i: x) os << (f++ ? ", " : ""), os << i; os << "}";
    return os;
}
 
void _print() {cerr << "]\n";}
 
template<typename T, typename... V>
void _print(T t, V... v) {cerr << t; if (sizeof...(v)) cerr <<", "; _print(v...);}
 
#ifdef DEBUG
#define dbg(x...) cerr << "\e[91m"<<__func__<<":"<<__LINE__<<" [" << #x << "] = ["; _print(x); cerr << "\e[39m" << endl;
#else
#define dbg(x...)
#endif

#define pii pair<int, int>
#define f first
#define s second

void solve(){
	int n;cin>>n;vector<array<int,3>>a(n);for(int i=0;i<n;i++)cin>>a[i][0]>>a[i][1]>>a[i][2];
	// bsearch on ans then we need to force some easy and some hard
	function<bool(int)> works=[&](int w){
		int easy=0,hard=0;
		// we want i total contests meaning we want i easy problems, which we will take from the contests sorted by how many easy we can take without taking away frmn the number of hard problems that is achievable
		int no_of_swap_options=0;
		for(auto i:a){
			int ha=min(i[1],i[0]);
			easy+=ha;
			if(ha<i[1]){
				int a=min(i[2],i[1]-ha);
				hard+=a;
				i[2]-=a;
				i[1]-=a;
			}
			// we are using everything in i[1]
			int x=min(i[1],i[2]);
			no_of_swap_options+=x;
		}
		if(easy>w){
			int x=min(easy-w,no_of_swap_options);
			hard+=x;
		}
		return min(easy,hard)>=w;
	};
	int ans=0;
	for(int i=60;i>=0;i--){
		if(works(ans+(1ll<<i)))ans+=(1ll<<i);
	}
	cout<<ans<<"\n";
}

int32_t main() {
	int t;cin>>t;while(t--)solve();
}

提出情報

ジャッジ結果