状压dp+二分

最新推荐文章于 2025-12-15 09:15:21 发布

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Problem Description

Finally, Little Q gets his weapon at level 10^5105 in the RPG game, now he is trying to beat the boss as soon as possible. The boss has HH units of health point (HP), Little Q needs to cause at least HH units of damage to beat the boss.

Little Q has learnt nn skills, labeled by 1,2,\dots,n1,2,…,n. Each skill can not be used multiple times, because there is not enough time for Little Q to wait for the skill to cool down. Assume Little Q uses the ii-th skill at the xx-th frame, the actor controlled by him will take t_iti frames to perform, which means Little Q will not be allowed to use other skills until the (x+t_i)(x+ti)-th frame. The length of the damage sequence of the ii-th skill is len_ileni, which means the skill will cause d_{i,j}di,j (0\leq j < len_i0≤j<leni) units of damage at the (x+j)(x+j)-th frame if Little Q uses the ii-th skill at the xx-th frame. Note that len_ileni can be greater than t_iti, for example, the burning skill can burn the boss for a long period, but takes a little time to cast the fire.

The game starts at the 00-th frame. Your task is to help Little Q beat the boss as soon as possible, or determine Little Q can't beat the boss using all the skills at most once.

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const int N=19,M=1e5+5;
ll f[1<<N];
ll d[N][M];
int t[N],len[N];
int p[1<<N];
int n;ll h;
bool check(int x)
{
	for(int i=0;i<(1<<n);i++)
	{
		int sum=0;
		f[i]=0;
		for(int j=0;j<n;j++)
		{	
			if((i>>j)&1) 
			{
				int now=p[i]-t[j+1];
				if(x-now+1>=1&&x-now+1<=len[j+1])
				f[i]=max(f[i],f[i^(1<<j)]+d[j+1][x-now+1]);
				else if(x-now+1>len[j+1])f[i]=max(f[i],f[i^(1<<j)]+d[j+1][len[j+1]]);
				else f[i]=max(f[i],f[i^(1<<j)]);
			}
		}
	}
	return (f[(1<<n)-1]>=h);
}
void solve()
{
	scanf("%d%lld",&n,&h);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&t[i],&len[i]);
		for(int j=1;j<=len[i];j++) 
		{
			scanf("%lld",&d[i][j]);
			d[i][j]+=d[i][j-1];
		}
	}
	for(int i=0;i<(1<<n);i++)
	{
		p[i]=0;
		for(int j=0;j<n;j++)
		{
			if((i>>j)&1) p[i]+=t[j+1];
		}
	}
	int l=0,r=2e6;
	int ans=-1;
	while(l<=r)
	{
		int m=(l+r)/2;
		if(check(m))
		{
			ans=m;
			r=m-1;
		}
		else l=m+1;
	}
	printf("%d\n",ans);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		solve();
	}