[XIN算法应用]NOI2020美食家

XIN(\(updated 2021.6.4\))

对于很多很多的题目，发现自己并不会之后，往往会直接冲上一个XIN队算法，然而，这样 \(\huge{\text{鲁莽}}\) 的行为只能获得 TLE，所以，我们要考虑如何拿到最大的部分分值。

noi 2020 美食家

题目

看完题目之后，发现这个题目的范围很鬼畜，似乎只能用 \(\mathcal O(log_2T)\) 的复杂的过去。。。。

之后大脑空白 \(1e9\) 分钟。。。。。。。。。。。。。。。。。

之后目光转向 1 ~ 8 的测试点，发现似乎可以用 \(\mathcal O(nT)\) 的复杂度拿到，所以开始考虑部分分数。

如果使用以往的XIN队算法，那么预测应该最多只拿到 4 个测试点，所以我们就要考虑非纯暴力解法

首先我们要推出方程：怎么推呢，用 \(\huge{\text{杠哥大定理}}\)。

所以方程就是:

\[f_{pos,time}=\max\limits_{link_{pos,to}}f_{to,time-edge[i].w} + c_i \]

之后就可以使用 \(\huge{\text{记忆化}}\)来解决

	int xin_team(int x,int tim)
	{
		if(f[x][tim]) return f[x][tim];
		if(x == 1 and !tim)	
			{ok = true;return c[x];}
		else if(tim < 0)
			return 0x7f7f7f7f7f7f7f7f;
		int maxx = -0x7f7f7f7f7f7f7f7f;
		for(register int i=0;i<link[x].size();++i)
		{
			register int y = link[x][i].pos;
			int zhuan = xin_team(y,tim - link[x][i].val);
			if(zhuan < 0x7f7f7f7f7f7f7f7f)
				maxx = max(maxx,zhuan);
		}
		for(register int i=1;i<=k;++i)
			if(x == a[i].x and tim == a[i].t)
				{f[x][tim] += a[i].y;break;}
		f[x][tim] += maxx + c[x];
//		cout<<"x = "<<x<<" time = "<<tim<<endl;
		return f[x][tim];
	}

这就是 核心代码

复杂度 \(\mathcal O(nT)\) 还算比较 \(\color{red}{\text{优秀}}\)

之后就可以拿到 \(40pts\) 了！

记录

\(code_{tot}:\)

#include<cmath>
#include<queue>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define int long long
#define m(c,num) memset(c,num,sizeof c)
#define INF 0x7f7f7f7f
#define debug cout<<"debug"<<endl
#define freopen eat = freopen
#define scanf a14 = scanf
namespace xin_io
{
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	char buf[1<<20],*p1 = buf,*p2 = buf; FILE *eat; int a14;
	inline void openfile() {freopen("t.txt","r",stdin);}
	inline void outfile()  {freopen("o.txt","w",stdout);}
	inline int get()
	{
		int s = 0,f = 1;register char ch = gc();
		while(!isdigit(ch))
		{if(ch == '-') f = -1;ch = gc();}
		while(isdigit(ch))
		{s = s * 10 + ch - '0';	ch = gc();}
		return  s * f;
	}
}
static const int maxn = 201,maxt = 52502;
using namespace xin_io;
namespace xin
{
	class xin_edge{public:int w,next,ver;}edge[maxn];
	class xin_data{public:int pos,val;xin_data(int x,int y):pos(x),val(y){}};
	class xin_food{public:int t,x,y;}a[maxn<<2];
	int head[maxn],zhi = 0;
	inline void add(int x,int y,int z)
	{edge[++zhi].ver = y;edge[zhi].w = z;edge[zhi].next = head[x]; head[x] = zhi;}
	int c[maxn];
	int n,m,t,k;
	bool ok = false;
	vector <xin_data> link[maxn];
	int f[maxn][maxt];
	int xin_team(int x,int tim)
	{
		if(f[x][tim]) return f[x][tim];
		if(x == 1 and !tim)	
			{ok = true;return c[x];}
		else if(tim < 0)
			return 0x7f7f7f7f7f7f7f7f;
		int maxx = -0x7f7f7f7f7f7f7f7f;
		for(register int i=0;i<link[x].size();++i)
		{
			register int y = link[x][i].pos;
			int zhuan = xin_team(y,tim - link[x][i].val);
			if(zhuan < 0x7f7f7f7f7f7f7f7f)
				maxx = max(maxx,zhuan);
		}
		for(register int i=1;i<=k;++i)
			if(x == a[i].x and tim == a[i].t)
				{f[x][tim] += a[i].y;break;}
		f[x][tim] += maxx + c[x];
//		cout<<"x = "<<x<<" time = "<<tim<<endl;
		return f[x][tim];
	}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get(); m = get(); t = get(); k = get();
//		memset(f,128,sizeof(f));
		if(t > 52501) {cout<<-1<<endl;return 0;}
		for(register int i=1;i<=n;++i) c[i] = get();
		for(register int i=1;i<=m;++i)
		{register int x = get(),y = get(),z = get();add(x,y,z),link[y].push_back(xin_data(x,z));}
		for(register int i=1;i<=k;++i) a[i].t = get(),a[i].x = get(),a[i].y = get();
		int ans = xin_team(1,t);
		if(ok)
			cout<<ans<<endl;
		else
			cout<<-1<<endl;
		return 0;
	}
}
signed main() {return xin::main();}