[考试总结]noip模拟6

我好菜啊

真上次第二这次倒二。。。

~~因为昨天还没有改完所有的题所以就留到今天来写博客了~~

这次考试总结的教训有很多吧，反正处处体现XIN某人的~~laji~~，自己考试的是后本以为一共四个题目，三个题目都没有看懂，然而考试结束以后才发现，自己是四个题目都没有看懂。~~cao~~

~~又成10分XIN了~~

不管了，菜就是菜了。

以后看到题目中不懂的玩意儿也不应该害怕，什么曼哈顿距离，自己看看样例就知道了，不应该弃掉的，并且在手推样例认为样例有锅的时候也应该返回去去看看题目，而不是一味地认为题目有锅。。。

$\huge{\text{菜}}$

~~来自 $\LaTeX$ 的侮辱~~

T1:

什么~~laji~~曼哈顿距离，反正对角线就完了。

。。。。。

然后记录每一个矩形，按照 $x$ 的大小进行排序，之后就计算相邻矩形形成的氢键数量就 $ok$ 了

~~为什么之考虑所有矩形不相连就有 $65pts$ 啊！！！~~

。。。。。

code：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<map>
using namespace std;
#define int long long
#define debug cout<<"debug"<<endl
#define freopen eat2 = freopen
#define scanf eat1 = scanf
namespace xin_io
{
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	#define scanf eat1 = scanf 
	#define freopen eat2 = freopen
	char buf[1<<20],*p1 = buf,*p2 = buf;FILE *eat2;
	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;
	}
}
using namespace xin_io; int eat1;
static const int maxn = 1e5+10;
#include<ctime>
#include<cstring>
#define mp make_pair
#define xd xin_data
#define rp(i,x,y) for(register int i=x;i<=y;++i)
#define bd(i,x,y) for(register int i=x;i>=y;--i)
namespace xin
{
	class xin_data
	{
		public:
			int x1,x2,y1,y2;
			friend bool operator < (xin_data x,xin_data y) {return x.x1 == y.x1 ? x.y1 < y.y1 : x.x1 < y.x1;}
	}p[maxn];
	int n,ans = 0;
	inline int find(int x,int y)
	{
		int tog;
		if(p[x].x2 + 1 == p[y].x1) 
		{
			if(p[y].y1 > p[x].y2 or p[y].y2 < p[x].y1) return 0;
			tog = abs(max(p[y].y2,p[x].y2) - min(p[y].y1,p[x].y1) + 1) - abs(p[y].y2 - p[x].y2) - abs(p[y].y1 - p[x].y1);
			if(p[x].y1 == p[y].y1 and p[x].y2 == p[y].y2) return tog * 2;
			if(p[x].y2 == p[y].y2 or p[x].y1 == p[y].y1) return tog * 2 - 1;
			return tog * 2;
		}
		if(p[x].y2 + 1 == p[y].y1)
		{
			if(p[y].x1 > p[x].x2 or p[x].x1 > p[y].x2) return 0;
			tog = abs(max(p[x].x2,p[y].x2) - min(p[x].x1,p[y].x1) + 1) - abs(p[x].x2 - p[y].x2) - abs(p[x].x1 - p[y].x1);
			if(p[x].x1 == p[y].x1 and p[x].x2 == p[y].x2) return tog * 2;
			if(p[x].x2 == p[y].x2 or p[x].x1 == p[y].x1) return tog * 2 - 1;
			return tog * 2;
		}
		return 0;
	}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get();
		rp(i,1,n) p[i].x1 = get(),p[i].y1 = get(),p[i].x2 = get(),p[i].y2 = get(),ans += (p[i].x2 - p[i].x1) * (p[i].y2 - p[i].y1) * 2;
		sort(p+1,p+n+1);
//		rp(i,1,n) cout<<"i = "<<i<<" p[i].x1 = "<<p[i].x1<<" p[i].y1 = "<<p[i].y1<<" p[i].x2 = "<<p[i].x2<<" p[i].y2 = "<<p[i].y2<<endl;
		rp(i,1,n)
			rp(j,i+1,n)
			{
				if(p[j].x1 - p[i].x2 > 1 ) break;
				if(min(p[i].x2,p[j].x2) - max(p[i].x1,p[j].x1) >= 0 and (p[i].y2 == p[j].y1 - 1 or p[j].y2 == p[i].y1 - 1))
				{
					if(p[i].y2 == p[j].y1 - 1)
					{
						ans += (min(p[i].x2,p[j].x2) - max(p[i].x1,p[j].x1)) * 2;
						if(p[i].x1 != p[j].x1) ans ++; if(p[i].x2 != p[j].x2) ans ++;
					}
					if(p[j].y2 == p[i].y1 - 1)
					{
						ans += (min(p[i].x2,p[j].x2) - max(p[i].x1,p[j].x1)) * 2;
						if(p[i].x1 != p[j].x1) ans ++; if(p[i].x2 != p[j].x2) ans ++;
					}
				}
				if(min(p[i].y2,p[j].y2) - max(p[i].y1 ,p[j].y1) >= 0 and (p[i].x2 == p[j].x1 - 1 or p[j].x2 == p[i].x1 - 1))
				{
					if(p[i].x2 == p[j].x1 - 1)
					{
						ans += (min(p[i].y2,p[j].y2) - max(p[i].y1,p[j].y1)) * 2;
						if(p[i].y1 != p[j].y1) ans ++; if(p[i].y2 != p[j].y2) ans ++;
					}
					if(p[j].x2 == p[i].x1 - 1)
					{
						ans += (min(p[i].y2,p[j].y2) - max(p[i].y1,p[j].y1)) * 2;
						if(p[i].y1 != p[j].y1) ans ++; if(p[i].y2 != p[j].y2) ans ++;
					}
				}
				if(p[i].x2 == p[j].x1 - 1 and (p[i].y2 == p[j].y1 - 1 or p[i].y1 == p[j].y2 + 1)) ans++;
				///cout<<"i = "<<i<<" ans = "<<ans<<endl;
			}
		cout<<ans<<endl;
		return 0;
	}
}
signed main() {return xin::main();}

T2:

$upd\;2021.6.13$

在经过~~二逼平衡树~~的洗礼之后，~~灵魂得到了升华~~然后这个题目的思路竟然也~~毛色吨开~~（~~laji~~谷歌输入法），然后道题目的正解也差不多了：

这个题目其实和二逼平衡树还是很像很像的，都是开很多的 $splay$ 而不是仅仅一棵，然后再进行统计或者是合并操作。。。

这个题目所要维护的东西还是比较新颖的，将时间作为排序的标准值。。。

然后就可以对于每一个点的信息进行插入，color time ，还有一些之前 $splay$ 套路的一些信息。

然后就是用 time 进行排序。。。

就这？？？

~~你维护试试，包准没入门~~

最后进行一遍 $dfs$，然后进行启发式合并就行了。。。。

简简单单

然后就是 $code$

#include<bits/stdc++.h>
using namespace std;
//#define int long long
#define debug cout<<"debug"<<endl
#define freopen eat2 = freopen
#define scanf eat1 = scanf
namespace xin_io
{
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	#define scanf eat1 = scanf 
	#define freopen eat2 = freopen
	char buf[1<<20],*p1 = buf,*p2 = buf;FILE *eat2;
	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;
	}
}
using namespace xin_io; int eat1;
static const int maxn = 2e6+10,mod = 1e6;
namespace xin
{
	class xin_splay{public:int cnt,col,tim,ch[2],val,fa,size;}t[maxn];
	class xin_edge{public:int next,ver;}edge[maxn]; int head[maxn],zhi = 0;
	int tot,ret;
	class xin_splay_tree
	{
		private:
			#define up(x) t[x].size = t[t[x].ch[1]].size + t[t[x].ch[0]].size + 1,t[x].cnt = t[t[x].ch[1]].cnt + t[t[x].ch[0]].cnt + t[x].val
 		public:
			int root;		  //to distinguish different splay trees
			map<int,int>vis;  //to record the tim of color
			inline int size() {return t[root].size;}
			inline void rotate(int x)
			{
				register int y = t[x].fa,z = t[y].fa;
				register int ki = t[y].ch[1] == x;
				t[z].ch[t[z].ch[1] == y] = x; t[x].fa = z;
				t[y].ch[ki] = t[x].ch[ki xor 1]; t[t[x].ch[ki xor 1]].fa = y;
				t[x].ch[ki xor 1] = y; t[y].fa = x;
				up(y);	up(x);
			}//checked  	
			inline void splay(int x,int goal)
			{
				while(t[x].fa != goal)
				{
					register int y = t[x].fa, z = t[y].fa;
					if(z != goal)
						(t[z].ch[0] == y) xor (t[y].ch[0] == x) ? rotate(x) : rotate(y);
					rotate(x);
				}
				up(x);
				root = x;
			}//checked
			inline void insert(int tim,int col,int val)
			{
				register int now = root,fa = 0;
				while(now and tim != t[now].tim) fa = now,now = t[now].ch[tim > t[now].tim];
				now = ++tot;
				if(fa) t[fa].ch[tim > t[fa].tim] = now;
				t[now].cnt = t[now].val = val;
				t[now].fa = fa; t[now].tim = tim; t[now].size = 1; t[now].col = col;
				splay(now,0);
			}//checked

			void change(int tim)
			{
				register int now = root;
				while(now and t[now].tim != tim)
					now = t[now].ch[tim > t[now].tim];
				if(now) t[now].val = 0;
				splay(now,0);
			}//checked
			inline bool judge(int tim,int col)
			{
				if(!vis[col]) {vis[col] = tim; return true;}
				else if(vis[col] > tim) {change(vis[col]);vis[col] = tim; return true;}
				else return false;
			}	//checked
			int findx(int x,int lim)
			{
				if(!x) return 0;
				if(t[t[x].ch[0]].size >= lim) findx(t[x].ch[0],lim);
				else if(t[t[x].ch[0]].size + 1 >= lim) return  ret + t[t[x].ch[0]].cnt + t[x].val;
				else ret += t[t[x].ch[0]].cnt + t[x].val,findx(t[x].ch[1],lim -t[t[x].ch[0]].size - 1);
			}//checked
			int findval(int lim)
			{
				ret = 0;
				if(!lim) return 0;
				if(lim >= t[root].size) return t[root].cnt;
				int ret = findx(root,lim); return ret;
			}//checked
	}a[maxn];
	int n,m,ans[maxn],k[maxn],rec[maxn]; //rec is to make a node for each splay tree
	inline void add(int x,int y) {edge[++zhi].ver = y;edge[zhi].next = head[x]; head[x] = zhi;}
	int now;
	inline void make(int x)
	{
		if(!x) return ;
		make(t[x].ch[0]);
		a[now].insert(t[x].tim,t[x].col,a[now].judge(t[x].tim,t[x].col));
		make(t[x].ch[1]);
	}
	void dfs(int x,int fa)
	{
		for(register int i=head[x];i;i=edge[i].next)
		{
			register int y = edge[i].ver;
			if(fa == y) continue;
			dfs(y,x);
			if(a[rec[x]].size() < a[rec[y]].size())
			{
			//	cout<<x<<endl;
				now = rec[y];
				make(a[rec[x]].root);
				rec[x] = now;
			}
			else 
			{
				now = rec[x];
				make(a[rec[y]].root);
			//	rec[y] = now;
			}
		}
		ans[x] = a[rec[x]].findval(k[x]);
	}
	inline double time() {return (double)clock()/(double)(CLOCKS_PER_SEC);}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get();
		for(register int i=1;i<=n-1;++i)
		{
			register int x = get(),y = get();
			add(x,y); add(y,x);
		}
		for(register int i=1;i<=n;++i) k[i] = get(),rec[i] = i;
		m = get();// time();
		for(register int i=1;i<=m;++i)
		{
			register int x = get(),col = get();
			a[rec[x]].insert(i,col,a[rec[x]].judge(i,col));
		}
		dfs(1,0);
		int q = get();
		while(q--)
			printf("%d\n",ans[get()]);
		return 0;
	}
}
signed main() {return xin::main();}

T3:

又是 $dark$ 佬。。。。

设 $f$ 表示概率，那么我们就可以知道：

\[f_i=f_{i-1}-sum_{i-1} \]

\[sum_{i-1} = \sum_{i=1}^{i-1} f_i \]

\[ans=\sum w_i * f_i \]

。。。。

所以$code:$



#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define int long long
#define debug cout<<"debug"<<endl
#define freopen eat2 = freopen
#define scanf eat1 = scanf
namespace xin_io
{
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	#define scanf eat1 = scanf 
	#define freopen eat2 = freopen
	char buf[1<<20],*p1 = buf,*p2 = buf;FILE *eat2;
	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;
	}
}
using namespace xin_io; int eat1;
static const int maxn = 1e5+10,mod = 1000000007;
#include<ctime>
#include<cstring>
namespace xin
{
	int n,m,k;
	int w[maxn];
	inline int ksm(int x,int y,int mod)
	{
		int ret = 1;
		while(y){if(y & 1) ret = ret * x % mod;x = x * x % mod; y >>= 1;}
		return ret;
	}
	inline int inv(int x) {return ksm(x,mod-2,mod);}
	int f[maxn],ans,s;
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get(); m = get(); k = get();
		int day = (n - k + 1) > 0 ? (n - k + 1) : 0;
		for(register int i=1;i<=m;++i) w[i] = get();//,w[i] = w[i-1] + w[i];
		for(register int i=1;i<=m;++i)
		{
			s = s + f[i-1] % mod;
			f[i] = (ksm(i,k,mod) * inv(ksm(m,k,mod)) - s) % mod;
			ans = (ans + w[i] * f[i]) % mod;
		}
		ans = ans * day % mod;
		cout<<ans<<endl;
		return 0;
	}
}
signed main() {return xin::main();}

T4:

暴搜。。。

只需要考虑每一个起始点和每一个结束点。。。

然后开始暴搜。。。

不需要建边，只要维护一个 $cost$ 数组就行了。。。

$cost$ 初始值就设置为 $0x7f7f7f7f7f$ 就行。。。

最后取得最小值。

一共 $n <= 12$ ，然而 $m$ 却$ <=2000$ ，重边很多。。。

之后就有 $70pts$ 了



#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define int long long
#define debug cout<<"debug"<<endl
#define freopen eat2 = freopen
#define scanf eat1 = scanf
namespace xin_io
{
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	#define scanf eat1 = scanf 
	#define freopen eat2 = freopen
	char buf[1<<20],*p1 = buf,*p2 = buf;FILE *eat2;
	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;
	}
}
using namespace xin_io; int eat1;
static const int maxn = 13;
#include<ctime>
#include<cstring>
#define m(c,num) memset(c,num,sizeof c)
namespace xin
{
	int n,m;
	int c[maxn][maxn];
	int temp;
	int num[maxn];
	int ans = 0x7f7f7f7f;
	void dfs(int now,int cnt)
	{
		if(cnt == n)
		{
			ans = min(ans,temp);
			return ;
		}
		if(temp >= ans) return ;
		for(register int i=1;i<=n;++i)
			if(!num[i])
				for(register int j=1;j<=n;++j)
					if(c[i][j] != 0x7f7f7f7f and i != j and num[j])
					{
						temp += num[j] * c[i][j]; num[i] = num[j] + 1;
						dfs(i,cnt+1);
						temp -= num[j] * c[i][j]; num[i] = 0;
					}
	}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get(); m = get(); for(register int i=1;i<=n;++i) for(register int j=0;j<=n;++j) c[i][j] = 0x7f7f7f7f;
		for(register int i=1;i<=m;++i)
		{
			register int x = get(),y = get();
			c[x][y] = c[y][x] = min(c[x][y],get()); 
		}
		for(register int i=1;i<=n;++i)
		{num[i] = 1; dfs(i,1); num[i] = 0;}
		cout<<ans<<endl;
		return 0;
	}
}
signed main() {return xin::main();}

之后考虑正解，实际上就是~~有枝就剪~~。

然后飞快。。。



#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
//#define int long long
#define debug cout<<"debug"<<endl
#define freopen eat2 = freopen
#define scanf eat1 = scanf
#define inf 0x7f7f7f7f
namespace xin_io
{
	#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
	#define scanf eat1 = scanf 
	#define freopen eat2 = freopen
	char buf[1<<20],*p1 = buf,*p2 = buf;FILE *eat2;
	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;
	}
}
using namespace xin_io; int eat1;
static const int maxn = 13;
#include<ctime>
#include<cstring>
#define m(c,num) memset(c,num,sizeof c)
namespace xin
{
	int n,m;
	int vis[maxn],rea[maxn][maxn],tot;
	int c[maxn][maxn],d[maxn],l[maxn];
	int temp,cnt;
	int ans = inf;
	inline void dfs(int num,int nd)
	{
		for(register int i=num;i<=cnt;++i)
		{
			if(tot + temp * l[vis[i]] >= ans) return;
			for(register int j=nd;j<=d[vis[i]];++j)
				if(!l[rea[vis[i]][j]])
				{
					vis[++cnt] = rea[vis[i]][j];
					temp -= c[vis[cnt]][rea[vis[cnt]][1]];
					tot += c[vis[i]][vis[cnt]] * l[vis[i]];
					l[vis[cnt]] = l[vis[i]] + 1;
					dfs(i,j+1);
					temp += c[vis[cnt]][rea[vis[cnt]][1]];
					tot -= c[vis[i]][vis[cnt]] * l[vis[i]];
					l[vis[cnt]] = 0;cnt--;
				}
				nd = 1;
		}
		if(cnt == n) {ans = min(ans,tot);return ;}
	}
	int pai;
	inline bool comp(int x,int y) {return c[pai][x] < c[pai][y];}
	inline short main()
	{
	#ifndef ONLINE_JUDGE
		openfile();
	#endif
		n = get(); m = get(); for(register int i=1;i<=n;++i) for(register int j=1;j<=n;++j) c[i][j] = inf; 
		for(register int i=1;i<=m;++i)
		{
			register int x = get(),y = get(),z = get();
			if(z <= c[x][y])
			{
				if(c[x][y] == inf) rea[x][++d[x]] = y,rea[y][++d[y]] = x;
				c[x][y] = c[y][x] = z;
			}
		}
		for(register int i=1;i<=n;++i)
		{	
			pai = i; sort(rea[i]+1,rea[i]+d[i]+1,comp);
			temp += c[i][rea[i][1]];
		}
		for(register int i=1;i<=n;++i)
		{
			cnt = 1; tot = 0;
			temp -= c[i][rea[i][1]];
			vis[1] = i;
			l[i] = 1;
			dfs(1,1);
			temp += c[i][rea[i][1]];
			l[i] = 0;
		}
		printf("%d\n",ans);
		return 0;
	}
}
signed main() {return xin::main();}

posted @ 2021-06-11 21:29 NP2Z 阅读(41) 评论(0) 编辑收藏举报

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[考试总结]noip模拟6

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