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题解 [AHOI2017/HNOI2017]大佬

作者:互联网

传送门

注意到题面里n很小,有\(n\leq100\)
考虑联系n的实际意义
n是你在大佬手中能活的天数
题面颇富深意

好了不闹了
n很小,对于\(40\%\)的数据,爆搜即可
考场上靠这个骗了40pts

对于满分做法
我是考完看了题解才开始写的
然而题解貌似写麻烦了
首先对大佬的伤害与特定日期无关,只与分配给蓄力/伤害的天数有关
所以先dp一下在保证不死的前提下最多能留出多少天输出
我们需要知道能否打出一个特定伤害
所以尝试爆搜一下,得到花费\(d\)时间可以打出的伤害\(f\)
回看题面,题面可转化为「对于每个\(c_i\),问是否能构造出一种花费\(d\)天的怼大佬方案,使伤害\(f\leqslant c_i\)且\(f+md-d\geqslant c_i\)」,这里md为最多输出天数
已知所有的\(f\)和\(d\),如何验证方案?
只怼一次或不怼\(O(n)\)扫一遍就好
怼两次,则需满足
(1)\(f_1+f_2 \leqslant c_i\)
(2)\(f_1+f_2+md-d_1-d_2 \geqslant c_i\)
(3)\(d_1+d_2 \leqslant md\)
将(2)移项,得\(f_1+f_2 \geqslant c_i+(d_1+d_2-md)\)
则满足(1)(2)必定同时满足(3)
仍然是区间取值问题,尝试将序列排序后固定一个端点
\(f_1, d_1, md\)固定,则check \(f_1+d_1+md+(f_2-d_2)_{max}\)是否\(\geqslant c_i\)即可
注意到这里\(f_1+f_2\)具有单调性,即(已由小到大排序过)对于\(f_i\)不可行的\(f_2\),对于\(f_{i+1}\)同样不可行
所以采用双指针,每个左端点继承上个左端点的右端点进行扫描
时间复杂度就优化到了\(O(n)\)

对于那部分爆搜:
直接爆搜时间复杂度显然不对,考虑进行剪枝
这里出现了我以前没应用过的剪枝方法:利用\(hash\)进行判重
显然爆搜时对于相同的参数,所得结果应该是一样的,否则请先调试
那么把整个参数表hash掉,可完成记忆化
不过在这道题里更显著的应用是,
对于一个相同的f,我们只关心最小用时d,所以可以再开个hash表存
根据题解,我们可以采用bfs而不是dfs,因为bfs满足最先搜到某个f时天数一定是最小的,可以利用此性质优化hash表

p.s. 这个双指针的思路着实清奇

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 110
#define MAXN 500010
#define ll long long 
#define ld long double
#define usd unsigned
#define ull unsigned long long
//#define int long long 

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
char buf[1<<21], *p1=buf, *p2=buf;
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m, mc;
int a[N], w[N], c[25], md, size, mblood, cnt;
int dp[N][N];

struct plan{int d, f; inline void build(int d_, int f_) {d=d_; f=f_;} plan(int d_, int f_):d(d_),f(f_){} plan(){}}met[MAXN];
inline bool operator == (plan a, plan b) {return a.f==b.f&&a.d==b.d;}
inline bool operator < (plan a, plan b) {return a.f<b.f;}
struct ele{int d, f, l; ele(){} ele(int d_, int f_, int l_):d(d_),f(f_),l(l_){} inline void build(int d_, int f_, int l_){d=d_; f=f_; l=l_;}};
inline bool operator == (ele a, ele b) {return a.f==b.f&&a.d==b.d&&a.l==b.l;}

struct hush_map1{
	static const int SIZE=101000;
	int size, head[SIZE];
	hush_map1():size(0){memset(head, 0, sizeof(head));}
	struct node{int dat, next;}e[SIZE*10];
	inline bool operator [] (int q) {
		ll t = q%SIZE;
		for (int i=head[t]; i; i=e[i].next)
			if (e[i].dat==q) return 1;
		node* k=&e[++size]; k->dat=q; k->next=head[t]; head[t]=size;
		return 0;
	}
}mp1;

struct hush_map2{
	static const int SIZE=1010000;
	int size, head[SIZE];
	hush_map2():size(0){memset(head, 0, sizeof(head));}
	struct node{ele p; int next;}e[SIZE*10];
	inline bool operator [] (ele q) {
		ll t = (1ll*(q.d<<7)*q.f+q.l)%SIZE;
		for (int i=head[t]; i; i=e[i].next)
			if (e[i].p==q) return 1;
		if (size<SIZE) {node* k=&e[++size]; k->p=q; k->next=head[t]; head[t]=size;}
		return 0;
	}
}mp2;

struct hush_c{
	static const int SIZE=1000;
	int size, head[SIZE];
	hush_c():size(0){memset(head, 0, sizeof(head));}
	struct node{int p; bool vis; int next;}e[SIZE<<1];
	inline void operator [] (int q) {
		int t=q%SIZE;
		for (int i=head[t]; i; i=e[i].next) 
			if (e[i].p==q) {
				if (!e[i].vis) e[i].vis=1,++cnt;
				break;
			}
	}
	void add(int q) {
		int t=q%SIZE;
		node* k=&e[++size]; k->p=q; k->vis=0; k->next=head[t]; head[t]=size;
	}
	bool query(int q) {
		int t=q%SIZE;
		for (int i=head[t]; i; i=e[i].next) 
			if (e[i].p==q)
				return e[i].vis;
		puts("error");
	}
}test;

void checkout() {for (int i=1; i<=m; ++i) printf("%d\n", test.query(c[i]));}

signed main()
{
	#ifdef DEBUG
	freopen("1.in", "r", stdin);
	#endif
	
	//cout<<double(sizeof(met)+sizeof(mp1.head)+sizeof(mp1.e))/1024/1024<<endl;
	
	n=read(); m=read(); mc=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	for (int i=1; i<=n; ++i) w[i]=read();
	for (int i=1; i<=m; ++i) c[i]=read(), mblood=max(mblood, c[i]), test.add(c[i]);
	
	for (int i=n; i; --i) 
		for (int j=a[i]; j<=mc; ++j) 
			dp[i][j] = max(max(dp[i+1][j-a[i]]+1, dp[i+1][j-a[i]+w[i]]), dp[i][j-1]); //, printf("dp[%d][%d]=%d\n", i, j, dp[i][j]);	
	md = dp[1][mc];
	//cout<<dp[1][mc]<<endl;
	for (int i=1; i<=m; ++i) if (c[i]<=md) test[c[i]];
	
	queue<ele> q;
	plan p; ele t, t2;
	q.push(ele(3, 1, 2));
	//met[++size].build(0, 0);
	while (!q.empty()) {
		t=q.front(); q.pop();
		//printf("%d %d\n", t.d, size);
		if (t.f>mblood) continue;
		
		if (t.f>1 && !mp1[t.f]) met[++size].build(t.d, t.f);
		//p.build(t.d, t.out+1);
		//if (!mp1[p]) met[++size]=p;
		
		if (t.d<md) {
			//printf("%d\n", mp2.size);
			#if 1
			t2.build(t.d+1, t.f, t.l+1);
			if (1ll*t.f*(t.l+1)<=mblood && !mp2[t2]) q.push(t2);
			t2.build(t.d+1, t.f*t.l, t.l);
			if (1ll*t.f*t.l<=mblood && !mp2[t2]) q.push(t2);
			#else
			if (1ll*t.f*(t.l+1)<=mblood) q.push(ele(t.d+1, t.f, t.l+1));
			if (t.f*t.l<=mblood) q.push(ele(t.d+1, t.f*t.l, t.l));
			#endif
		}
	}
	for (int i=1; i<=m; ++i) for (int j=1; j<=size; ++j) if (met[j].f<=c[i] && met[j].f+md-met[j].d>=c[i]) test[c[i]];
	//for (int i=1; i<=size; ++i) cout<<met[i].d<<' '<<met[i].f<<endl;
	//cout<<size<<endl<<endl;
	#if 0
	for (int i=1,t; i<=size; ++i) 
		for (int j=md-met[i].d; j>=0; --j) {
			t = met[i].f+j;
			if (t) test[t], cout<<t<<endl;
		}
	if (cnt>=m) {checkout(); return 0;}
	#endif
	
	sort(met+1, met+size+1);
	for (int i=1,ci,val; i<=m; ++i) {
		ci = c[i];
		if (test.query(ci)) continue;
		plan *l=met+1, *r=met+size;
		while (l<=r) {
			val = md+l->f-l->d;
			//cout<<l-met<<' '<<r-met<<endl;
			while (l->f+r->f > ci) --r;
			//cout<<(val+r->f-r->d)<<endl;
			if (val+r->f-r->d >= ci) {test[ci]; /*cout<<"pos1"<<endl;*/ goto jump1;}
			++l;
		}
		jump1: ;
	}
	checkout();

	return 0;
}

标签:md,head,int,题解,long,HNOI2017,AHOI2017,define,size
来源: https://www.cnblogs.com/narration/p/14866746.html