Why is HN's style so weird this year?
D1T1
I think this is an interesting question
If the operator before the ith number is and, this bit is set to 1, otherwise it is 0, and the resulting binary number is recorded as xx.
Considering each bit separately, for the i-th bit, if the j-th number is 1, then this bit is set to 1, otherwise it is 0, and the resulting binary number is recorded as $b_i$.
Taking the left as the lowest bit, it is easy to prove by prefix induction that the result of the i-th bit is 1 if and only if $x < b_i$
We sort b from large to small, and the result is set to c, then the answer is not zero only if in the order of c, there is no 0 in r before 1.
Find the position of the first 0 in r, assuming k, then the solution x must satisfy $c_k≤x<c_{k–1}$, so the answer is $c_{k – 1} – c_k$
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=5000+7;
const ll mod=1e9+7;
int n,m,q,p[maxn],id[maxn];
ll mi[maxn],num[maxn];
bool G[maxn][maxn];
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
bool cmp(const int a,const int b) {
Rep(i,n,1) {
if(G[a][i]<G[b][i]) return 1;
if(G[a][i]>G[b][i]) return 0;
}
}
int main() {
freopen("hunt.in","r",stdin);
freopen("hunt.out","w",stdout);
read(n); read(m); read(q);
int x;
mi[0]=1;
For(i,1,n) {
mi[i]=mi[i-1]*2%mod;
For(j,1,m) {
scanf("%1d",&x);
if(x==1) {
G[j][i]=1;
num[j]=(num[j]+mi[i-1])%mod;
}
}
}
For(i,1,m) p[i]=i;
sort(p+1,p+m+1,cmp);
For(i,1,m) id[p[i]]=i;
p[m+1]=m+1; num[m+1]=mi[n];
int l,r;
For(i,1,q) {
l=0; r=m+1;
For(j,1,m) {
scanf("%1d",&x);
if(x==1) r=min(r,id[j]);
else l=max(l,id[j]);
}
if(l>r) printf("0\n");
else printf("%lld\n",(num[p[r]]-num[p[l]]+mod)%mod);
}
return 0;
}
D1T2
场上我一直在想，把环剖成链，倍长，线段树上第i个位置维护$T_i + 2 \times n - i$
于是我死在不知道怎么维护线段树上了，因为我假设$i$是最后取的点，那么我需要$i-n+1$到$i$的最大值。
但是如果在线段树上第$i$个位置维护$p_i =T_i - i$，那么对于一个位置$i$，假设$i$是第一个取的点，那么我们需要$i+1$到$2 \times n$的最大值就可以了。
对于线段树上的一个点所对应区间[l,r]，我们维护一个$p_i$的最大值和$\min \limits_{i \in [l,mid]}\{i+ \max \limits_{i \le j \le r}p_j\}$
维护的方法和bzoj2957楼房重建这道题一样。
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=2e5+7,INF=1e8;
int n,m,O,a[maxn],lastans;
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
int maxnum[4*maxn],p[4*maxn],qpos,qx;
int find(int pos,int l,int r,int x) {
if(x>=maxnum[pos]) return x+l+n;
if(l==r) return p[pos];
int mid=(l+r)>>1,rs=INF;
rs=min(rs,find(pos<<1|1,mid+1,r,x));
if(x>maxnum[pos<<1|1]) rs=min(rs,find(pos<<1,l,mid,x));
else rs=min(rs,p[pos<<1]);
return rs;
}
void bld(int pos,int l,int r) {
if(l==r) {
p[pos]=a[l]+l+n;
maxnum[pos]=a[l];
return;
}
int mid=(l+r)>>1;
bld(pos<<1,l,mid);
bld(pos<<1|1,mid+1,r);
maxnum[pos]=max(maxnum[pos<<1],maxnum[pos<<1|1]);
p[pos]=min(p[pos<<1|1],lastans=p[pos<<1]=find(pos<<1,l,mid,maxnum[pos<<1|1]));
}
void chge(int pos,int l,int r) {
if(l==r) {
p[pos]=qx+l+n;
maxnum[pos]=qx;
return;
}
int mid=(l+r)>>1;
if(qpos<=mid) chge(pos<<1,l,mid);
else chge(pos<<1|1,mid+1,r);
maxnum[pos]=max(maxnum[pos<<1],maxnum[pos<<1|1]);
p[pos]=min(p[pos<<1|1],lastans=p[pos<<1]=find(pos<<1,l,mid,maxnum[pos<<1|1]));
}
int main() {
freopen("circle.in","r",stdin);
freopen("circle.out","w",stdout);
read(n); read(m); read(O);
For(i,1,n) {
read(a[i]);
a[i+n]=a[i];
a[i]-=i;
a[i+n]-=i+n;
}
bld(1,1,2*n);
int x,y;
printf("%d\n",--lastans);
For(i,1,m) {
read(x); read(y);
if(O) x^=lastans,y^=lastans;
qpos=x; qx=y-x;
chge(1,1,2*n);
qpos=x+n; qx=y-x-n;
chge(1,1,2*n);
printf("%d\n",--lastans);
}
return 0;
}
D1T3不会
D2T1
这是啥子题哟，骗分？
把带锁的门看作有向边，其他部分可以缩点，然后可以按照拓扑序做。
场上写了一个带log的st表二分，后来证明了一下复杂度。
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=1e6+7,maxt=23,W=20,INF=0x3f3f3f3f;
int n,m,Q,a[maxn],last[maxn],next[maxn],st[maxn][maxt];
int ld[maxn],rd[maxn];
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
int get_rch(int p,int rs) {
Rep(i,W,0) if(st[rs][i]>=p) rs+=(1<<i);
return rs;
}
int main() {
freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
read(n); read(m); read(Q);
int x,y;
For(i,1,m) {
read(x); read(y);
a[x]=y;
}
For(i,1,n) if(!a[i]) a[i]=INF;
x=1;
For(i,1,n) {
last[i]=x;
if(a[i]!=INF&&a[i]<=i) x=i+1;
}
x=n;
Rep(i,n,1) {
next[i]=x;
if(a[i-1]!=INF&&a[i-1]>=i) x=i-1;
}
For(i,1,n) st[i][0]=a[i];
For(r,1,W) For(i,1,n-(1<<r)+1)
st[i][r]=min(st[i][r-1],st[i+(1<<r-1)][r-1]);
For(i,1,n) {
ld[i]=rd[i]=i;
if(a[i-1]==INF) { //both side road
ld[i]=ld[i-1];
rd[i]=rd[i-1];
continue;
}
rd[i]=min(next[i],get_rch(i,i));//ef
if(last[i]==i||rd[i]<a[i-1]) continue;//cannot reach i-1
while(ld[i]>last[i]||rd[i]<next[i]) {
if(ld[i]>last[i]&&a[ld[i]-1]<=rd[i]) ld[i]=ld[ld[i]-1];
else if(rd[i]<next[i]&&a[rd[i]]>=ld[i]) //include a[rd[i]]==INF
rd[i]=min(next[i],get_rch(ld[i],rd[i]+1));
else break;
}
}
For(i,1,Q) {
read(x); read(y);
if(y>=ld[x]&&y<=rd[x]) printf("YES\n");
else printf("NO\n");
}
// cerr<<clock()<<"\n";
return 0;
}
D2T2
明显先建树，然后直接根据平均数贪心跟父亲合并，用STL里面的set或者priority_queue。发现priority_queue的push快，但是pop慢。
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<set>
using namespace std;
#define ll long long
#define db long double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=1e6+7;
const db eps=1e-7;
ll n,fa[maxn],w[maxn],root;db sum[maxn],p[maxn];
int R[maxn],NXT[maxn];
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
int dcmp(db x) {return fabs(x)<eps? 0:(x>0? 1:-1);}
int f[maxn];
int find(int x){return x==f[x]? x:f[x]=find(f[x]);}
int fir[maxn],nxt[maxn],to[maxn],e=0;
void add(int x,int y) {
to[++e]=y;nxt[e]=fir[x];fir[x]=e;
}
bool vis[maxn];
inline bool s(int pos) {
if(vis[pos]) return 1;
vis[pos]=1;
for(int y=fir[pos];y;y=nxt[y])
if(s(to[y])) return 1;
return 0;
}
struct Node{
int pos;db w;
Node(int pos,db w):pos(pos),w(w){}
bool operator < (const Node& b) const{return w<b.w;}
};
multiset<Node>G;
multiset<Node>::iterator it;
ll solve() {
if(s(0)) return -1;
For(i,1,n) if(!vis[i]) return -1;
For(i,1,n) G.insert(Node(i,sum[i]=w[i]));
int x,y;db now;
For(i,1,n) {
do {
it=G.begin();
x=it->pos; now=it->w;
G.erase(it);
}while(find(x)!=x||dcmp(now-sum[x]/p[x]));//
y=find(fa[x]);
sum[y]+=sum[x];
p[y]+=p[x];
f[x]=y;
NXT[R[y]]=x; R[y]=R[x];
if(y) G.insert(Node(y,sum[y]/p[y]));
}
// cerr<<clock()<<"\n";
int pos=0; ll rs=0;
For(i,1,n) {
pos=NXT[pos];
rs+=w[pos]*(ll)i;
}
return rs;
}
int main() {
freopen("perm.in","r",stdin);
freopen("perm.out","w",stdout);
read(n);
For(i,1,n) p[i]=1,f[i]=R[i]=i;
For(i,1,n) read(fa[i]),add(fa[i],i);
For(i,1,n) read(w[i]);
printf("%lld\n",solve());
return 0;
}
D2T3
dp水题，但是要爆int/卡空间
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define lc son[pos][0]
#define rc son[pos][1]
const int maxn=40000+3,maxt=41;
ll n,A[maxn],B[maxn],C[maxn],son[maxn][2];
map<int,ll>dp[maxn];
char cc; ll ff;
template<typename T>void read(T& aa) {
aa=0;cc=getchar();ff=1;
while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
if(cc=='-') ff=-1,cc=getchar();
while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
aa*=ff;
}
int T(int x,int y) {return x*100+y;}
void s(int pos,int l1,int l2) {
if(pos>=n) {
For(i,0,l1) For(j,0,l2) dp[pos][T(i,j)]=C[pos]*(A[pos]+i)*(B[pos]+j);
return;
}
s(lc,l1+1,l2); s(rc,l1,l2+1);
For(i,0,l1) For(j,0,l2)
dp[pos][T(i,j)]=
min(dp[lc][T(i,j)] + dp[rc][T(i,j+1)],
dp[lc][T(i+1,j)]+ dp[rc][T(i,j)]);
}
int main() {
freopen("road.in","r",stdin);
freopen("road.out","w",stdout);
read(n); int x,y;
For(i,1,n-1) {
read(x); read(y);
if(x<0) x=n-1-x;
if(y<0) y=n-1-y;
son[i][0]=x; son[i][1]=y;
}
For(i,1,n) {
read(A[i+n-1]);
read(B[i+n-1]);
read(C[i+n-1]);
}
s(1,0,0);
printf("%lld\n",dp[1][0]);
return 0;
}