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!!!!!
}