AIM Tech Round 4 Div. 1

时间：2021-02-23 16:40:30

　　A：显然最优方案是对所形成的置换的每个循环排个序。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;

#define ll long long

#define N 100010

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}

int gcd(int n,int m){return m==0?n:gcd(m,n%m);}

int read()

{

	int x=0,f=1;char c=getchar();

	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}

	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();

	return x*f;

}

int n,a[N],c[N];

vector<int>  b;

bool flag[N];

signed main()

{

#ifndef ONLINE_JUDGE

	freopen("a.in","r",stdin);

	freopen("a.out","w",stdout);

#endif

	n=read();

	for (int i=1;i<=n;i++) c[i]=a[i]=read();

	sort(c+1,c+n+1);

	for (int i=1;i<=n;i++) a[i]=lower_bound(c+1,c+n+1,a[i])-c;

	int ans=0;

	for (int i=1;i<=n;i++)

	if (!flag[i])

	{

		flag[i]=1;int x=i;

		while (!flag[a[x]]) x=a[x],flag[x]=1;

		ans++;

	}

	cout<<ans<<endl;

	memset(flag,0,sizeof(flag));

	for (int i=1;i<=n;i++)

	if (!flag[i])

	{

		b.clear();

		int cnt=1;flag[i]=1;int x=i;b.push_back(i);

		while (!flag[a[x]]) x=a[x],flag[x]=1,cnt++,b.push_back(x);

		printf("%d ",cnt);

		for (int j=0;j<cnt;j++) printf("%d ",b[j]);printf("\n");

	}

	return 0;

	//NOTICE LONG LONG!!!!!

}

　　B：随机问1000个位置，然后找到x在其中哪个区间内暴力询问即可。注意最好不要rand。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<ctime>

#include<vector>

#include<chrono>

#include<random>

using namespace std;

#define ll long long

#define N 50010

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}

int gcd(int n,int m){return m==0?n:gcd(m,n%m);}

int read()

{

	int x=0,f=1;char c=getchar();

	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}

	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();

	return x*f;

}

int n,start,x,val[N],nxt[N];

bool flag[N];

struct data{int p,val,nxt;

    bool operator <(const data&a) const

    {

        return val<a.val;

    }

}a[N];

signed main()

{

#ifndef ONLINE_JUDGE

	freopen("a.in","r",stdin);

	freopen("a.out","w",stdout);

#endif

	n=read(),start=read(),x=read();

	mt19937 rnd(time(0));

	cout<<"?"<<' '<<start<<endl;

	flag[start]=1;cin>>val[start]>>nxt[start];

	if (val[start]>=x) {cout<<"!"<<' '<<val[start]<<endl;return 0;}

	for (int i=1;i<=999;i++)

	{

		int x=rnd()%n+1;

		cout<<"?"<<' '<<x<<endl;

		flag[x]=1;cin>>val[x]>>nxt[x];

	}

	int t=0;

	for (int i=1;i<=n;i++) if (flag[i]) a[++t].p=i,a[t].nxt=nxt[i],a[t].val=val[i];

	sort(a+1,a+t+1);

	a[t+1].val=1010000000;

	for (int i=1;i<=t;i++)

	if (a[i].val<=x&&a[i+1].val>x)

	{

	    int u=a[i].nxt,v=a[i].val;

	    if (v>=x) {cout<<"!"<<' '<<v<<endl;return 0;}

	    while (u!=-1&&v<x)

	    {

	        cout<<"?"<<' '<<u<<endl;

	        cin>>v>>u;

	    }

	    if (v>=x)

	    {

	        cout<<"!"<<' '<<v<<endl;return 0;

	    }

	    else {cout<<"!"<<' '<<-1<<endl;return 0;}

	}

	return 0;

	//NOTICE LONG LONG!!!!!

}

　　C：容易发现重心的各棵子树的点集是不能改变的。然后一堆人比如我就扔个点分上去肯定就假了。事实上可以通过这种操作将任意一棵子树展开成链，链可以再转化成菊花，分别消耗低于n次。第一步逐个考虑当前点的每个子树（都已经形成链），将当前点与父亲切断，并将该子树所形成的链的底端接到父亲。第二步由链底端自下往上切掉，并将原本的菊花中心接上即可。注意判一下有两个重心的情况。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<set>

#include<cassert>

using namespace std;

#define ll long long

#define N 400010

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}

int gcd(int n,int m){return m==0?n:gcd(m,n%m);}

int read()

{

	int x=0,f=1;char c=getchar();

	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}

	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();

	return x*f;

}

int n,p[N],size[N],ansx[N],ansy[N],ansz[N],fa[N],top[N],pre[N],u,root,root2,t;

set<int> e[N];

struct data{int to,nxt;

}edge[N<<1];

void addedge(int x,int y){t++;e[x].insert(y);edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}

void make(int k,int from)

{

    size[k]=1;

    for (int i=p[k];i;i=edge[i].nxt)

    if (edge[i].to!=from)

    {

        make(edge[i].to,k);

        size[k]+=size[edge[i].to];

    }

}

int findroot(int k,int s,int from)

{

    int mx=0;

    for (int i=p[k];i;i=edge[i].nxt)

    if (edge[i].to!=from&&size[edge[i].to]>size[mx]) mx=edge[i].to;

    if ((size[mx]<<1)>s) return findroot(mx,s,k);

    else return k;

}

void solve(int k,int from)

{

	fa[k]=from;

	for (int i=p[k];i;i=edge[i].nxt)

	if (edge[i].to!=from&&edge[i].to!=root&&edge[i].to!=root2) solve(edge[i].to,k);

	for (int i=p[k];i;i=edge[i].nxt)

	if (edge[i].to!=from&&edge[i].to!=root&&edge[i].to!=root2)

	{

		if (!top[k]) top[k]=edge[i].to;

		ansx[++u]=fa[k];

		ansy[u]=k;

		ansz[u]=edge[i].to;

		e[ansx[u]].erase(ansy[u]),e[ansx[u]].insert(ansz[u]);

		e[ansy[u]].erase(ansx[u]),e[ansz[u]].insert(ansx[u]);

		fa[k]=top[edge[i].to];

	}

	if (!top[k]) top[k]=k;

}//����kΪ�����������һ����

int get(int x)

{

	while (1)

	{

		auto it=e[x].begin();

		while (it!=e[x].end()&&(*it)==pre[x]) it++;

		if (it==e[x].end()) return x;

		else pre[*it]=x,x=*it;

	}

}

signed main()

{

#ifndef ONLINE_JUDGE

	freopen("a.in","r",stdin);

	freopen("a.out","w",stdout);

#endif

	n=read();

	for (int i=1;i<n;i++)

	{

		int x=read(),y=read();

		addedge(x,y),addedge(y,x);

	}

	make(1,1);

	root=findroot(1,n,1);

	if (n%2==0)

	{

		for (int i=1;i<=n;i++)

		if (size[i]==n/2) {root2=i;break;}

	}

	if (root2)

	{

		for (int i=p[root2];i;i=edge[i].nxt)

		if (edge[i].to!=root)

		{

			solve(edge[i].to,root2);

			pre[top[edge[i].to]]=root2;

			int x=get(top[edge[i].to]),k=x;

			while (k!=root2)

			{

				ansx[++u]=pre[k];

				ansy[u]=k;

				ansz[u]=x;

				k=pre[k];

			}

		}

	}

	for (int i=p[root];i;i=edge[i].nxt)

	if (edge[i].to!=root2)

	{

		solve(edge[i].to,root);

		pre[top[edge[i].to]]=root;

		int x=get(top[edge[i].to]),k=x;

		while (k!=root)

		{

			ansx[++u]=pre[k];

			ansy[u]=k;

			ansz[u]=x;

			k=pre[k];

		}

	}

	printf("%d\n",u);

	for (int i=1;i<=u;i++)  printf("%d %d %d\n",ansx[i],ansy[i],ansz[i]);

	return 0;

	//NOTICE LONG LONG!!!!!

}

　　D：注意到dij的堆如果可以用桶替代就能做到线性最短路，这要求值域较小，而题面保证了总边权变化量不超过1e6。考虑怎么仅对变化量做最短路。一开始跑完dij后，将每条边(u,v)的边权重赋值为disu+w(u,v)-disv建成新图，每次修改就在新图中进行修改并跑桶优化dij，再将得到的最短路值累加，并继续对新图进行重赋值即可。好像每次边权只+1并没有什么卵用？

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<queue>

#include<vector>

using namespace std;

#define ll long long

#define N 100010

#define inf 10000000000000000ll

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}

int gcd(int n,int m){return m==0?n:gcd(m,n%m);}

int read()

{

	int x=0,f=1;char c=getchar();

	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}

	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();

	return x*f;

}

int n,m,Q,p[N],t;

ll d[N],d2[N];

bool flag[N];

vector<int> a[N];

struct data{int to,nxt;ll len;

}edge[N];

void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}

struct data2

{

	int x;ll d;

	bool operator <( const data2&a) const

	{

		return d>a.d;

	}

};

priority_queue<data2> q;

void dijkstra()

{

	memset(d,42,sizeof(d));d[1]=0;

	q.push((data2){1,0});

	for (;;)

	{

		while (!q.empty()&&flag[q.top().x]) q.pop();

		if (q.empty()) break;

		data2 x=q.top();q.pop();

		flag[x.x]=1;

		for (int i=p[x.x];i;i=edge[i].nxt)

		if (x.d+edge[i].len<d[edge[i].to])

		{

			d[edge[i].to]=x.d+edge[i].len;

			q.push((data2){edge[i].to,d[edge[i].to]});

		}

	}

	for (int i=1;i<=n;i++)

		for (int j=p[i];j;j=edge[j].nxt)

		edge[j].len=d[i]+edge[j].len-d[edge[j].to];

}

void update(int c)

{

	memset(d2,42,sizeof(d2));

	memset(flag,0,sizeof(flag));

	for (int i=0;i<=c;i++) a[i].clear();

	a[0].push_back(1);d2[1]=0;

	for (int i=0;i<=c;i++)

	{

		for (int k=0;k<a[i].size();k++)

		{

			int x=a[i][k];

			if (!flag[x])

			{

				flag[x]=1;

				for (int j=p[x];j;j=edge[j].nxt)

				if (d2[x]+edge[j].len<d2[edge[j].to]&&d2[x]+edge[j].len<=c)

				{

					d2[edge[j].to]=d2[x]+edge[j].len;

					a[d2[edge[j].to]].push_back(edge[j].to);

				}

			}

		}

	}

	for (int i=1;i<=n;i++)

		for (int j=p[i];j;j=edge[j].nxt)

		edge[j].len=d2[i]+edge[j].len-d2[edge[j].to];

	for (int i=1;i<=n;i++) if (d2[i]<inf) d[i]+=d2[i];

}

signed main()

{

#ifndef ONLINE_JUDGE

	freopen("a.in","r",stdin);

	freopen("a.out","w",stdout);

#endif

	n=read(),m=read(),Q=read();

	for (int i=1;i<=m;i++)

	{

		int x=read(),y=read(),z=read();

		addedge(x,y,z);

	}

	dijkstra();

	for (int i=1;i<=Q;i++)

	{

		int op=read();

		if (op==1)

		{

			int x=read();

			if (d[x]>inf) printf("-1\n");

			else printf("%I64d\n",d[x]);

		}

		else

		{

			int c=read();

			for (int j=1;j<=c;j++) edge[read()].len++;

			update(c);

		}

	}

	return 0;

	//NOTICE LONG LONG!!!!!

}

　　

相关文章