题解 ZROI3
T1
与《滑动窗口》类似，用单调队列维护
#include
#include
#include
#include
#include
using namespace std;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
#define ror(a,b,c) for(register int a=c;a>=b;--a)
#define tor(a,b) for(register int a=head[b];a;a=nxt[a])
const int MAX=1e5+5;
int n,m,input,num[MAX];
struct data{
int val,pos;
};
deque line;
inline int read();
int main(){
#ifndef ONLINE_JUDGE
freopen("test.in","r",stdin);
// freopen("test.out","w",stdout);
#endif
m=read();
while(input=read(),input!=-1){
num[++n]=input;
}
lor(i,1,n){
while(!line.empty()&&line.front().pos=m) printf("%d\n",line.front().val);
}
return 0;
}
inline int read(){
char tmp=getchar(); int sum=0; bool flag=false;
while(tmp<'0'||tmp>'9'){
if(tmp=='-') flag=true;
tmp=getchar();
}
while(tmp>='0'&&tmp<='9'){
sum=(sum<<1)+(sum<<3)+tmp-'0';
tmp=getchar();
}
return flag?-sum:sum;
}
T2
看起来是树链剖分..但其实用不着。由于只有”子树“相关的操作，因此把树打成DFS序后修改连续的区间即可
#include
#include
#include
#include
#include
using namespace std;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
#define ror(a,b,c) for(register int a=c;a>=b;--a)
#define tor(a,b) for(register int a=head[b];a;a=nxt[a])
const int MAX=1e5+5;
int n,m; char input[MAX];
int root=1,ecnt,edge[MAX<<1],head[MAX],nxt[MAX<<1];
int hei[MAX],size[MAX],fa[MAX],ind[MAX],rev[MAX];
int val[MAX<<2],sum[MAX<<2],lazy_rev[MAX<<2];
inline int read();
inline void insert(int,int,int);
void dfs(int,int);
void build(int,int,int);
int print(int,int,int,int);
void change(int,int,int,bool);
void pushdown(int,int,int);
void modify_rev(int,int,int,int,int);
int query(int,int,int,int,int);
int main(){
#ifndef ONLINE_JUDGE
freopen("test.in","r",stdin);
#endif
n=read(); m=read(); cin>>input+1;
lor(i,1,n-1){
int u=read(),v=read();
insert(u,v,++ecnt); insert(v,u,++ecnt);
}
dfs(root,root);
build(1,1,n);
lor(i,1,m){
char type; int x; cin>>type; x=read();
switch(type){
case 'S':
modify_rev(1,1,n,rev[x],rev[x]+size[x]-1);
break;
case 'Q':
printf("%d\n",query(1,1,n,rev[x],rev[x]+size[x]-1));
break;
}
}
return 0;
}
inline int read(){
char tmp=getchar(); int sum=0; bool flag=false;
while(tmp<'0'||tmp>'9'){
if(tmp=='-') flag=true;
tmp=getchar();
}
while(tmp>='0'&&tmp<='9'){
sum=(sum<<1)+(sum<<3)+tmp-'0';
tmp=getchar();
}
return flag?-sum:sum;
}
inline void insert(int from,int to,int id){
nxt[id]=head[from]; head[from]=id; edge[id]=to;
}
void dfs(int u,int f){
hei[u]=hei[f]+1;
fa[u]=f;
ind[++ind[0]]=u;
rev[u]=ind[0];
tor(i,u){
int v=edge[i]; if(v==f) continue;
dfs(v,u);
size[u]+=size[v];
}
size[u]++;
}
void build(int p,int l,int r){
if(l==r) {val[p]=input[ind[l]]-'0'; sum[p]=val[p]==1; return;}
int mid=(l+r)>>1;
build(p<<1,l,mid); build(p<<1|1,mid+1,r);
sum[p]=sum[p<<1]+sum[p<<1|1];
}
int print(int p,int l,int r,int k){
if(l==k&&k==r) return val[p];
pushdown(p,l,r);
int mid=(l+r)>>1;
if(k<=mid) return print(p<<1,l,mid,k);
if(mid+1<=k) return print(p<<1|1,mid+1,r,k);
}
void change(int p,int l,int r,bool rev){
lazy_rev[p]^=rev;
if(rev) sum[p]=(r-l+1)-sum[p];
if(l==r){
if(lazy_rev[p]) val[p]^=1;
lazy_rev[p]=false;
}
}
void pushdown(int p,int l,int r){
if(!lazy_rev[p]) return;
int mid=(l+r)>>1;
change(p<<1,l,mid,lazy_rev[p]);
change(p<<1|1,mid+1,r,lazy_rev[p]);
lazy_rev[p]=false;
}
void modify_rev(int p,int l,int r,int L,int R){
if(L<=l&&r<=R) {change(p,l,r,true); return;}
pushdown(p,l,r);
int mid=(l+r)>>1;
if(L<=mid) modify_rev(p<<1,l,mid,L,R);
if(mid+1<=R) modify_rev(p<<1|1,mid+1,r,L,R);
sum[p]=sum[p<<1]+sum[p<<1|1];
}
int query(int p,int l,int r,int L,int R){
if(L<=l&&r<=R) return sum[p];
pushdown(p,l,r);
int mid=(l+r)>>1,ans=0;
if(L<=mid) ans+=query(p<<1,l,mid,L,R);
if(mid+1<=R) ans+=query(p<<1|1,mid+1,r,L,R);
return ans;
}
T3
在考场上就意识到了和GSS4极其相似，但败在了数学证明上..
结论:任意数字\(x\)经过一次有效的取模之后（模数小于 x），其大小必定小于\(\frac{x}{2}\)
证明：若 \(mod\geq \frac{x}{2}\),则\(x-mod\leq \frac {x}{2}\)；若\(mod < \frac{x}{2}\),则\(x
这样就可以保证任意节点的操作次数不大于\(log_2 10^9=30\)，时间复杂度得到保证。其余细节与GSS4大同小异
#include
#include
#include
#include
using namespace std;
typedef long long ll;
#define lor(a,b,c) for(register int a=b;a<=c;++a)
const int MAX=1e5+5;
int n,m; ll init[MAX];
ll val_max[MAX<<2],val_sum[MAX<<2];
inline int read();
void build(int,int,int);
void modify_mod(int,int,int,int,int,ll);
void modify_as(int,int,int,int,ll);
ll query(int,int,int,int,int);
int main(){
#ifndef ONLINE_JUDGE
freopen("test.in","r",stdin);
#endif
n=read(); m=read();
lor(i,1,n) scanf("%lld",&init[i]);
build(1,1,n);
lor(i,1,m){
int type,l,r; ll x; type=read();
switch(type){
case 1:
l=read(); r=read();
printf("%lld\n",query(1,1,n,l,r));
break;
case 2:
l=read(); r=read(); scanf("%lld",&x);
modify_mod(1,1,n,l,r,x);
break;
case 3:
l=read(); scanf("%lld",&x);
modify_as(1,1,n,l,x);
break;
}
}
return 0;
}
inline int read(){
char tmp=getchar(); int sum=0; bool flag=false;
while(tmp<'0'||tmp>'9'){
if(tmp=='-') flag=true;
tmp=getchar();
}
while(tmp>='0'&&tmp<='9'){
sum=(sum<<1)+(sum<<3)+tmp-'0';
tmp=getchar();
}
return flag?-sum:sum;
}
void build(int p,int l,int r){
if(l==r) {val_sum[p]=val_max[p]=init[l]; return;}
int mid=(l+r)>>1;
build(p<<1,l,mid); build(p<<1|1,mid+1,r);
val_sum[p]=val_sum[p<<1]+val_sum[p<<1|1];
val_max[p]=max(val_max[p<<1],val_max[p<<1|1]);
}
void modify_mod(int p,int l,int r,int L,int R,ll k){
if(l==r) {val_sum[p]%=k; val_max[p]=val_sum[p]; return;}
int mid=(l+r)>>1;
if(L<=mid&&val_max[p<<1]>=k) modify_mod(p<<1,l,mid,L,R,k);
if(mid+1<=R&&val_max[p<<1|1]>=k) modify_mod(p<<1|1,mid+1,r,L,R,k);
val_sum[p]=val_sum[p<<1]+val_sum[p<<1|1];
val_max[p]=max(val_max[p<<1],val_max[p<<1|1]);
}
void modify_as(int p,int l,int r,int pos,ll k){
if(l==pos&&pos==r) {val_sum[p]=k; val_max[p]=k; return;}
int mid=(l+r)>>1;
if(pos<=mid) modify_as(p<<1,l,mid,pos,k);
if(mid+1<=pos) modify_as(p<<1|1,mid+1,r,pos,k);
val_sum[p]=val_sum[p<<1]+val_sum[p<<1|1];
val_max[p]=max(val_max[p<<1],val_max[p<<1|1]);
}
ll query(int p,int l,int r,int L,int R){
if(L<=l&&r<=R) return val_sum[p];
int mid=(l+r)>>1; ll ans=0;
if(L<=mid) ans+=query(p<<1,l,mid,L,R);
if(mid+1<=R) ans+=query(p<<1|1,mid+1,r,L,R);
return ans;
}