The 22nd Japanese Olympiad in Informatics (JOI 2022/2023) Final Round 题解

交题：https://cms.ioi-jp.org/documentation

A

给一个序列 $a_1,\cdots ,a_n$。
执行$n$个操作，第$i$个操作为找出第$i$个数前离其最近且与它相同的数的位置，把这两个数之间的数全部赋值$a_i$。求最后的序列。

考虑第$i$个操作执行完后，$i$之前每个数一定是连续出现正好一段或不出现。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \For(j,m-1) cout<<a[i][j]<<' ';\cout<<a[i][m]<<endl; \} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{int x=0,f=1; char ch=getchar();while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}return x*f;
} 
int a[201000];
map<int,pair<int,int> > h;
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);int n=read();For(i,n) a[i]=read();	stack <pair<pair<int,int> , int > > st;For(i,n) {if(st.empty() || !h.count(a[i]) || h[a[i]] ==mp(0,0) ) {st.push(mp(mp(i,i),a[i]));h[a[i]]=mp(i,i);}else {while(!st.empty()) {auto p=st.top();st.pop();if(p.se==a[i]) {h[p.se]=mp(p.fi.fi,i);st.push(mp(h[p.se],a[i]));break;}else {h[p.se]=mp(0,0);}}}}while(!st.empty()) {auto p=st.top();st.pop();Fork(i,p.fi.fi,p.fi.se) a[i]=p.se;}For(i,n) cout<<a[i]<<endl;return 0;
}

B

给$n$个点对，每个点对$(x,y)$可以覆盖$S=(a,b)|b<=y,|a-x|<=y-b$。问取多少个点对能覆盖所有点对。

经典题

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \For(j,m-1) cout<<a[i][j]<<' ';\cout<<a[i][m]<<endl; \} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{int x=0,f=1; char ch=getchar();while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}return x*f;
} 
int n;
vector<pair<int,int> > v;
int main()
{
//	freopen("B.in","r",stdin);
//	freopen(".out","w",stdout);int n=read();For(i,n) {int a=read(),b=read();v.pb(mp(b-a,a+b));}sort(ALL(v));stack<pair<int,int> > st;	for(int i=0;i<n;i++) {auto now=v[i];while(!st.empty()){auto t=st.top();if(t.fi<=now.fi && t.se <=now.se) {st.pop(); }else break;}st.push(now);}cout<<SI(st)<<endl;return 0;
}

C

考虑n*n的四连通矩阵，每次可以上下左右走一个。
格子上有颜色（黑、白），且只有白色能走。现在你希望令2个白色格子连通。一次操作为把$n\ast n$的矩阵赋值为白色。问至少几次操作实现目标。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \For(j,m-1) cout<<a[i][j]<<' ';\cout<<a[i][m]<<endl; \} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{int x=0,f=1; char ch=getchar();while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}return x*f;
} 
int r,c,n,sx,sy,tx,ty;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
bool inside(int x,int y) {return 1<=x && x<=r &&1<=y &&y<=c;
}
vector<string> a;
bool state(int x,int y){return a[x-1][y-1]=='#';
}
pair<int,int> dis[6000000+10];
int id(int x,int y) {return c*(x-1)+y;
}
void pri(pair<int,int> p) {printf("(%d,%d)",p.fi,p.se);
}
void pri(vector<pair<int,int> > v) {for(auto a:v) {pri(a);cout<<":";int x=a.fi,y=a.se;pri(dis[id(x,y)]);cout<<" ";}cout<<endl;
}
void bfs() {vector<pair<int,int> > q0,qa,qb;int nowdis=0;For(i,r*c) dis[i]=mp(INF,INF);dis[id(sx,sy)]=mp(0,n);q0.pb(mp(sx,sy));while(SI(q0)) {int nxdis=nowdis+1;// relax q0for(int i=0;i<SI(q0);i++) {auto t=q0[i];int x=t.fi,y=t.se;auto now_dis=dis[id(x,y)];Rep(di,4) {int nx=x+dir[di][0],ny=y+dir[di][1];if(!inside(nx,ny)) continue;if(dis[id(nx,ny)]!=mp(INF,INF)) continue;int sta=state(nx,ny);if(sta==0) { //whitedis[id(nx,ny)] = now_dis;q0.pb(mp(nx,ny));}}}	// q0 -> qaRep(i,SI(q0)) {qa.pb(q0[i]);}Rep(i,SI(qa)) {auto t=qa[i];int x=t.fi,y=t.se;auto now_dis=dis[id(x,y)];Rep(di,2) {int nx=x+dir[di][0],ny=y+dir[di][1];if(!inside(nx,ny)) continue;if(dis[id(nx,ny)]!=mp(INF,INF)) continue;int sta=state(nx,ny);if(now_dis.fi==nowdis) {dis[id(nx,ny)] = mp(nxdis,1);}else if(now_dis.se+1<=n){dis[id(nx,ny)] = mp(nxdis,now_dis.se+1);}else continue;qa.pb(mp(nx,ny));}}//qa -> abRep(i,SI(qa)) {qb.pb(qa[i]);}Rep(i,SI(qb)) {auto t=qb[i];int x=t.fi,y=t.se;auto now_dis=dis[id(x,y)];Fork(di,2,3) {int nx=x+dir[di][0],ny=y+dir[di][1];if(!inside(nx,ny)) continue;if(dis[id(nx,ny)]!=mp(INF,INF)) continue;int sta=state(nx,ny);if(now_dis.fi==nowdis) {dis[id(nx,ny)]=mp(nxdis,n+1);}else if(now_dis.fi==nxdis && now_dis.se<n &&  n+1<=2*n ) {dis[id(nx,ny)]=mp(nxdis,n+1);}else if(now_dis.fi==nxdis && now_dis.se==n && n+2<=2*n ) {dis[id(nx,ny)]=mp(nxdis,n+2);}else if(now_dis.fi==nxdis && now_dis.se>n && now_dis.se+1<=2*n ) {dis[id(nx,ny)]=mp(nxdis,now_dis.se+1);}else continue;qb.pb(mp(nx,ny));}}
//		
//		cout<<nowdis<<endl;
//		cout<<"q0"<<endl;
//		pri(q0);
//		cout<<"qa"<<endl;
//		pri(qa);
//		cout<<"qb"<<endl;
//		pri(qb);
//		//ab -> qc(q0)q0.resize(0);for(int i=0;i<qb.size();i++) {auto t=qb[i];int x=t.fi,y=t.se;auto now_dis=dis[id(x,y)];if(now_dis.fi==nxdis)q0.pb(qb[i]);}qa.resize(0),qb.resize(0);nowdis++;}
//	For(i,r) {
//		For(j,c) pri(dis[id(i,j)]),putchar(' ');
//		puts("");
//	}cout<<dis[id(tx,ty)].fi<<endl;}
int main()
{
//	freopen("C.in","r",stdin);
//	freopen(".out","w",stdout);cin>>r>>c>>n;
//	For(i,r) For(j,c) cout<<id(i,j)<<' ';cin>>sx>>sy>>tx>>ty;For(i,r) {string s;cin>>s;a.pb(s);}bfs();return 0;
}

D Cat Exercise

给一个$n$个节点的树，点权$a_i$。
执行如下操作：

1. 选取点权最大的点
2. 删除这个点及其相连的边，若有剩余连通块中取一个，跳回1。否则结束。

问操作1取的点权的和最大值。