hzwer分块九题(暂时持续更新)
hzwer分块9题
分块1:区间加法,单点查询
Code
#include<bits/stdc++.h>
#define in(i) (i=read())
using namespace std;
const int N=5e5+10,inf=2e9;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+(i^48),i=getchar();
return ans*f;
}
int n,m,block;
int pos[N],v[N],tag[N];
void add(int a,int b,int c) {
for(int i=a;i<=min(pos[a]*block,b);i++) v[i]+=c;
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++) v[i]+=c;
for(int i=pos[a]+1;i<=pos[b]-1;i++) tag[i]+=c;
}
int main()
{
in(n); block=sqrt(n);
for(int i=1;i<=n;i++) in(v[i]);
for(int i=1;i<=n;i++)
pos[i]=(i-1)/block+1;
for(int i=1;i<=n;i++) {
int op,a,b,c;
in(op),in(a),in(b),in(c);
if(!op) add(a,b,c);
else cout<<v[b]+tag[pos[b]]<<endl;//a,c没用
}
}
分块2:区间加法,求区间内小于x的个数
提示:暴力+二分,但是由于一些数加了之后会改变顺序,所以要重构
Code
#include<bits/stdc++.h>
#define in(i) (i=read())
using namespace std;
const int N=5e5+10,inf=2e9;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+(i^48),i=getchar();
return ans*f;
}
int n,m,block;
int pos[N],v[N],tag[N];
vector<int>t[N];
void rebuild(int x) {
t[x].clear();
for(int i=(x-1)*block+1;i<=min(n,x*block);i++)
t[x].push_back(v[i]);
sort(t[x].begin(),t[x].end());
}
void add(int a,int b,int c) {
for(int i=a;i<=min(pos[a]*block,b);i++) v[i]+=c;
rebuild(pos[a]);
if(pos[a]!=pos[b]) {
for(int i=(pos[b]-1)*block+1;i<=b;i++) v[i]+=c;
rebuild(pos[b]);
}
for(int i=pos[a]+1;i<=pos[b]-1;i++) tag[i]+=c;
}
int query(int a,int b,int c,int ans=0) {
for(int i=a;i<=min(pos[a]*block,b);i++)
if(v[i]+tag[pos[a]]<c) ans++;
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++)
if(v[i]+tag[pos[b]]<c) ans++;
for(int i=pos[a]+1;i<=pos[b]-1;i++) {
int x=c-tag[i];
ans+=lower_bound(t[i].begin(),t[i].end(),x)-t[i].begin();
}return ans;
}
int main()
{
in(n); block=sqrt(n);
for(int i=1;i<=n;i++) in(v[i]);
for(int i=1;i<=n;i++) {
pos[i]=(i-1)/block+1;
t[pos[i]].push_back(v[i]);
}
for(int i=1;i<=pos[n];i++) sort(t[i].begin(),t[i].end());
for(int i=1;i<=n;i++) {
int op,a,b,c;
in(op),in(a),in(b),in(c);
if(!op) add(a,b,c);
else cout<<query(a,b,c*c)<<endl;
}
}
分块3:区间加法,求前驱
提示:暴力+二分(set)
Code
#include<bits/stdc++.h>
#define in(i) (i=read())
using namespace std;
const int N=5e5+10,inf=2e9;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+(i^48),i=getchar();
return ans*f;
}
int n,m,block;
int pos[N],v[N],tag[N];
set<int>s[N];
void add(int a,int b,int c) {
for(int i=a;i<=min(pos[a]*block,b);i++) {
s[pos[a]].erase(v[i]);
s[pos[a]].insert(v[i]+=c);
}
if(pos[a]!=pos[b]) {
for(int i=(pos[b]-1)*block+1;i<=b;i++) {
s[pos[b]].erase(v[i]);
s[pos[b]].insert(v[i]+=c);
}
}
for(int i=pos[a]+1;i<=pos[b]-1;i++) tag[i]+=c;
}
int query(int a,int b,int c,int ans=-1) {
for(int i=a;i<=min(pos[a]*block,b);i++)
if(v[i]+tag[pos[a]]<c) ans=max(ans,v[i]+tag[pos[a]]);
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++)
if(v[i]+tag[pos[b]]<c) ans=max(ans,v[i]+tag[pos[b]]);
for(int i=pos[a]+1;i<=pos[b]-1;i++) {
int x=c-tag[i];
set<int>::iterator it =s[i].lower_bound(x);
if(it==s[i].begin()) continue; it--;
ans=max(ans,*it+tag[i]);
}return ans;
}
int main()
{
in(n); block=sqrt(n);
for(int i=1;i<=n;i++) in(v[i]);
for(int i=1;i<=n;i++) {
pos[i]=(i-1)/block+1;
s[pos[i]].insert(v[i]);
}
for(int i=1;i<=n;i++) {
int op,a,b,c;
in(op),in(a),in(b),in(c);
if(!op) add(a,b,c);
else cout<<query(a,b,c)<<endl;
}
}
分块4:区间加法,区间求和
Code
#include<bits/stdc++.h>
#define in(i) (i=read())
using namespace std;
const int N=5e5+10,inf=2e9;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+(i^48),i=getchar();
return ans*f;
}
int n,m,block;
int pos[N],v[N],tag[N],sum[N];
void add(int a,int b,int c) {
for(int i=a;i<=min(pos[a]*block,b);i++) v[i]+=c,sum[pos[a]]+=c;
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++) v[i]+=c,sum[pos[b]]+=c;
for(int i=pos[a]+1;i<=pos[b]-1;i++) tag[i]+=c;
}
int query(int a,int b,int ans=0) {
for(int i=a;i<=min(pos[a]*block,b);i++)
ans+=v[i]+tag[pos[a]];
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++)
ans+=v[i]+tag[pos[b]];
for(int i=pos[a]+1;i<=pos[b]-1;i++)
ans+=sum[i]+block*tag[i];
return ans;
}
int main()
{
in(n); block=sqrt(n);
for(int i=1;i<=n;i++) in(v[i]);
for(int i=1;i<=n;i++)
pos[i]=(i-1)/block+1,sum[pos[i]]+=v[i];
for(int i=1;i<=n;i++) {
int op,a,b,c;
in(op),in(a),in(b),in(c);
if(!op) add(a,b,c);
else printf("%d\n",query(a,b)%(c+1));
}
}
分块5:区间开方,区间求和
提示:因为开方一些次数后,数会变成0/1,我们对不全为0/1的块暴力开方,否则跳过
Code
#include<bits/stdc++.h>
#define in(i) (i=read())
using namespace std;
const int N=5e5+10,inf=2e9;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+(i^48),i=getchar();
return ans*f;
}
int n,m,block;
int pos[N],v[N],tag[N],sum[N],vis[N];
void solve(int x) {
if(vis[x]) return;
vis[x]=1; sum[x]=0;
for(int i=(x-1)*block+1;i<=x*block;i++) {
v[i]=sqrt(v[i]); sum[x]+=v[i];
if(v[i]>1) vis[x]=0;
}
}
void update(int a,int b,int c) {
for(int i=a;i<=min(b,pos[a]*block);i++) {
sum[pos[a]]-=v[i];
v[i]=sqrt(v[i]);
sum[pos[a]]+=v[i];
}
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++) {
sum[pos[b]]-=v[i];
v[i]=sqrt(v[i]);
sum[pos[b]]+=v[i];
}
for(int i=pos[a]+1;i<=pos[b]-1;i++) solve(i);
}
int query(int a,int b,int ans=0) {
for(int i=a;i<=min(pos[a]*block,b);i++)
ans+=v[i]+tag[pos[a]];
if(pos[a]!=pos[b])
for(int i=(pos[b]-1)*block+1;i<=b;i++)
ans+=v[i]+tag[pos[b]];
for(int i=pos[a]+1;i<=pos[b]-1;i++)
ans+=sum[i]+block*tag[i];
return ans;
}
int main()
{
in(n); block=sqrt(n);
for(int i=1;i<=n;i++) in(v[i]);
for(int i=1;i<=n;i++)
pos[i]=(i-1)/block+1,sum[pos[i]]+=v[i];
for(int i=1;i<=n;i++) {
int op,a,b,c;
in(op),in(a),in(b),in(c);
if(!op) update(a,b,c);
else printf("%d\n",query(a,b)%(c+1));
}
}
分块6:插入值,单点查询
提示:暴力插入,然后把后面的节点全部后移,当块的大小较大后重构
Code
#include<bits/stdc++.h>
#define in(i) (i=read())
#define pr pair<int,int>
using namespace std;
const int N=5e5+10,inf=2e9;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') ans=(ans<<1)+(ans<<3)+(i^48),i=getchar();
return ans*f;
}
int n,m,blo,cnt;
int v[N],st[N];
vector<int>t[N];
pr find(int a) {//找到是哪个块
int x=1;
while(a>t[x].size())
a-=t[x].size(),x++;
return make_pair(x,a-1);//因为vector为0~n-1
}
void rebuild() {
cnt=0;
for(int i=1;i<=m;i++) {
for(vector<int>::iterator it=t[i].begin();it!=t[i].end();it++)
st[++cnt]=*it;
t[i].clear();
}
blo=sqrt(cnt);
for(int i=1;i<=cnt;i++)
t[(i-1)/blo+1].push_back(st[i]);
m=(cnt-1)/blo+1;
}
void insert(int a,int b) {
pr q=find(a);
t[q.first].insert(t[q.first].begin()+q.second,b);
if(t[q.first].size()>10*blo) rebuild();//重构
}
int main()
{
in(n); blo=sqrt(n);
for(int i=1;i<=n;i++) in(v[i]);
for(int i=1;i<=n;i++)
t[(i-1)/blo+1].push_back(v[i]);
m=(n-1)/blo+1;//最终块是哪个,会更新
for(int i=1;i<=n;i++) {
int op,a,b,c;
in(op),in(a),in(b),in(c);
if(!op) insert(a,b);
else {
pr q=find(b);
printf("%d\n",t[q.first][q.second]);
}
}
}
博主蒟蒻,随意转载.但必须附上原文链接
http://www.cnblogs.com/real-l/
博主蒟蒻,随意转载.但必须附上原文链接
http://www.cnblogs.com/real-l/
http://www.cnblogs.com/real-l/
