AT_abc317_d [ABC317D] President 题解

考虑每一个 $x_i < y_i$ 的组，拿回 $z_i$ 的票需要 $\lceil \dfrac{y_i-x_i}{2}\rceil$ 个人从 $y_i$ 变成 $x_i$。

直接算出两个人的得分 $a,b$，则至少需要 $\lceil \dfrac{b-a}{2}\rceil$ 的票。

直接背包就好了。

#pragma GCC optimize("-Ofast,fast-math,-inline")
#include <bits/stdc++.h>
using namespace std;
#define int long long

const int N = 1e5 + 5, MOD = 1e9 + 7; // Remember to change

int n;

struct Node
{
	long long x, y, z;
	Node(long long _x, long long _y, long long _z): x(_x), y(_y), z(_z){}
	Node(){}
}a[N];

namespace FastIo
{
	#define QUICKCIN ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
	int read()
	{
		char ch = getchar();
		int x = 0, f = 1;
		while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
		while (ch == '-')
		{
			f = -f;
			ch = getchar();
		}
		while (ch >= '0' && ch <= '9')
		{
			x = (x << 1) + (x << 3) + (ch ^ 48);
			ch = getchar();
		}
		return x * f;
	}
	template<class T>
	void write(T x)
	{
		if (x < 0)
		{
			putchar('-');
			x = -x;
		}
		if (x > 9) write(x / 10);
		putchar(x % 10 + '0');
	}
	template<class T>
	void writeln(T x)
	{
		write(x);
		putchar('\n');
	}
}

int dp[N];
vector<pair<int, int> > v;

signed main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	cin>>n;
	int s1=0,s2=0;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i].x>>a[i].y>>a[i].z;
		if(a[i].x<a[i].y)
		{
			v.emplace_back(make_pair((a[i].y-a[i].x+1)/2, a[i].z));
			s2+=a[i].z;
		}
		else s1+=a[i].z;
	}
	if(s1>s2) cout<<"0\n";
	else
	{
		int p=(s2-s1+1)/2;
		//cout<<p<<"\n";
		memset(dp,0x3f,sizeof dp);
		dp[0]=0;
		for(auto&i:v)
		{
			//cout<<i.first<<" " << i.second << "\n";
			for(int j=100000;j>=i.second;j--)
			{
				dp[j]=min(dp[j],dp[j-i.second]+i.first);
			}
		}
		int ans=1e18;
		for(int i=p;i<=100000;i++) ans=min(ans, dp[i]);
		cout<<ans<<"\n";
	}
	return 0;
}