NKOJ P2300 what the boss moving bricks
I do not change the segment tree format handstand Women
KONO 题面 哒!
Summer vacation, you go to what his boss, what the boss asks you what would you say would only move bricks.
So, what the boss wanted to test the strength of what you move bricks:
what the boss has n blocks of white tiles (numbered 1 to n), he told you to put this number by n block became a straight line.
Do you think put in line to get away with it? What the boss is a very strange man, he kept to your whims, his command has the following three:
. A first to a second color c b brick into the brick.
II. The first to a second color for the bricks b c d for the brick into the brick color.
III. Replying to a b-th bricks in which there are several different colors.
You have to quickly give the correct answer every question, or you would not get the job!
(What the boss of brick with up to 7 different colors, numbered 1-7, white numbered 1).
Data range:
$ n-\ 200000 Leq, m \ $ 100000 Leq
Ten of eleven segment tree explode all around because the interval assigned the wrong value Pushdown, I split the whole person
At first glance only seven colors that this is like pressure. But I think for a long time did not want to come out like how to write press, then staged a violent approach. Open seven segment tree, record the number of each color bricks within the current range.
Each No. 1 To erase all the values on the remaining six trees, tree to the current interval to interval length the bricks.
Each operation to find No. 2 \ (C \) tree in the target range memory section bricks, these sections erased, then \ (D \) tree these intervals to the interval length.
No. 3 on each operating statistics for each tree target range of memory there is no value.
Very violent, but after doing a little tiny Feifei Feifei optimized to run faster.
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#define tp(p) if((p<<1|1)<=1200010)
using namespace std;
char *p1,*p2,buf[1<<20];
#define GC (p1==p2&&(p1=buf,p2=buf+fread(buf,1,1<<20,stdin),p1==p2)?0:(*(p1++)))
//#define GC getchar()
inline int in()
{
int x=0,w=0;
char ch=0;
while(!isdigit(ch)){
w|=ch=='-';
ch=GC;
}
while(isdigit(ch)){
x=(x<<3)+(x<<1)+(ch^48);
ch=GC;
}
return w?-x:x;
}
int n,m;
inline int ls(int x){return x<<1;}
inline int rs(int x){return x<<1|1;}
const int maxn=300010;
struct tree{
int sum[4*maxn],tag[4*maxn];
void pushup(int p){
sum[p]=sum[ls(p)]+sum[rs(p)];
}
void pushdown(int p,int l,int r){
int mid=((l+r)>>1);
if(tag[p]==1){
sum[ls(p)]=mid-l+1;
sum[rs(p)]=r-mid;
tag[ls(p)]=tag[rs(p)]=1;
}
else if(tag[p]==-1){
sum[ls(p)]=sum[rs(p)]=0;
tag[rs(p)]=tag[ls(p)]=-1;
}
tag[p]=0;
}
void ins(int l,int r,int p,int l1,int r1,int k)
{
pushdown(p,l1,r1);
if(l<=l1&&r>=r1){
if(k==1){
sum[p]=r1-l1+1;
tag[p]=1;
}
else{
sum[p]=0;
tag[p]=-1;
}
return;
}
int mid=((l1+r1)>>1);
if(mid>=l)ins(l,r,ls(p),l1,mid,k);
if(mid<r)ins(l,r,rs(p),mid+1,r1,k);
pushup(p);
return;
}
bool gs(int l,int r,int p,int l1,int r1)
{
pushdown(p,l1,r1);
if(l<=l1&&r>=r1){
return sum[p];
}
int mid=((l1+r1)>>1);
bool res=0;
if(mid>=l)res=(res|gs(l,r,ls(p),l1,mid));
if(mid<r)res=(res|gs(l,r,rs(p),mid+1,r1));
return res;
}
}tr[8];
void change(int l,int r,int p,int l1,int r1,int c,int d)
{
tr[c].pushdown(p,l1,r1);
tr[d].pushdown(p,l1,r1);
if(tr[c].sum[p]==0)return;
if(l<=l1&&r>=r1&&tr[c].sum[p]==r1-l1+1){
tr[c].sum[p]=0;
tr[c].tag[p]=-1;
tr[d].sum[p]=r1-l1+1;
tr[d].tag[p]=1;
return;
}
if(l1==r1)return;
int mid=((l1+r1)>>1);
if(mid>=l)change(l,r,ls(p),l1,mid,c,d);
if(mid<r)change(l,r,rs(p),mid+1,r1,c,d);
tr[c].pushup(p);
tr[d].pushup(p);
return;
}
int main()
{
int i,j;
n=in();m=in();
tr[1].ins(1,n,1,1,n,1);
for(i=1;i<=m;i++){
int ch,a,b,c;
ch=in();a=in();b=in();
switch(ch){
case 1:{
c=in();
for(j=1;j<=7;j++){
if(j==c)continue;
tr[j].ins(a,b,1,1,n,0);
}
tr[c].ins(a,b,1,1,n,1);
break;
}
case 2:{
int d;
c=in();d=in();
change(a,b,1,1,n,c,d);
break;
}
case 3:{
for(j=1;j<=7;j++)
{
if(tr[j].gs(a,b,1,1,n)){
printf("%d ",j);
}
}
printf("\n");
break;
}
}
}
return 0;
}Next is God zmr guy like Pressure practices
#include<stdio.h>
#include<bits/stdc++.h>
#define cut(l,r) ((l+r)>>1)
#define lson(i) (i<<1)
#define rson(i) ((i<<1)|1)
#define ll long long
#define intinf 1000000000
#define llinf 1000000000000000000LL
#define reint register int
#define st first
#define nd second
using namespace std;
char *p1,*p2,buf[1<<20];
#define GC (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?0:*p1++)
//#define GC getchar()
inline int rd()
{
int w=0,x=0;char ch=0;
while(!isdigit(ch)){w|=ch=='-';ch=GC;}
while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=GC;}
return w?-x:x;
}
char pbuf[1<<20],*pp=pbuf;
void push(const char c){
if(pp-pbuf==(1<<20)) fwrite(pbuf,1,(1<<20),stdout),pp=pbuf;
*pp++=c;
}
inline void wt(int x){
static int sta[35];int top=0;
do{sta[top++]=x%10,x/=10;}while(x);
while(top) push(sta[--top]+'0');
}
#define pn push('\n')
#define pk push(' ')
const int maxn=2e5+110;
const int maxr=maxn*4;
int n,m,a,b,c,d,id;
typedef pair<bool,pair<short,short> > arr;
struct node{
int l,r;
short sum;
vector<arr>to_do;
}tree[maxn*4];
inline void push_up(int i)
{
tree[i].sum=(tree[lson(i)].sum|tree[rson(i)].sum);
}
inline void push_down(int i)
{
//标记为1表示换成颜色为change的砖
//标记为2表示将change_from颜色的砖转换成颜色为change_to的砖
if(!tree[i].to_do.size())return;
arr tmp,t;
int ls=lson(i),rs=rson(i);
short fr,to;
vector<arr>::iterator it;
for(it=tree[i].to_do.begin();it!=tree[i].to_do.end();it++)
{
tmp=*it;fr=tmp.nd.st;to=tmp.nd.nd;
if(tmp.st==false)
{
while(tree[ls].to_do.size()) tree[ls].to_do.erase(tree[ls].to_do.begin());
while(tree[rs].to_do.size()) tree[rs].to_do.erase(tree[rs].to_do.begin());
tree[ls].sum=(1<<(fr-1));
tree[rs].sum=(1<<(fr-1));
t.st=false;t.nd.st=fr;
tree[ls].to_do.push_back(t);
tree[rs].to_do.push_back(t);
}
else
{
t.st=true;t.nd.st=fr;t.nd.nd=to;
if(tree[ls].sum&(1<<(fr-1)))
{
tree[ls].sum=(tree[ls].sum^(1<<fr-1));
tree[ls].sum=(tree[ls].sum|(1<<(to-1)));
tree[ls].to_do.push_back(t);
}
if(tree[rs].sum&(1<<(fr-1)))
{
tree[rs].sum=(tree[rs].sum^(1<<fr-1));
tree[rs].sum=(tree[rs].sum|(1<<(to-1)));
tree[rs].to_do.push_back(t);
}
}
}
while(tree[i].to_do.size()) tree[i].to_do.erase(tree[i].to_do.begin());
}
inline void build(int i,int l,int r)
{
tree[i].l=l;tree[i].r=r;tree[i].sum=1;
if(l==r) return;
int mid=cut(tree[i].l,tree[i].r);
build(lson(i),l,mid);
build(rson(i),mid+1,r);
push_up(i);
}
inline void change1(int i,int l,int r,int col)
{
if(tree[i].l>=l&&tree[i].r<=r)
{
while(tree[i].to_do.size()) tree[i].to_do.erase(tree[i].to_do.begin());
arr tmp;tmp.st=false;tmp.nd.st=col;
tree[i].to_do.push_back(tmp);
tree[i].sum=(1<<(col-1));
return;
}
push_down(i);
int mid=cut(tree[i].l,tree[i].r);
if(mid>=l) change1(lson(i),l,r,col);
if(mid<r) change1(rson(i),l,r,col);
push_up(i);
}
inline void change2(int i,int l,int r,int fr,int to)
{
if(tree[i].l>=l&&tree[i].r<=r)
{
arr tmp;tmp.st=true;tmp.nd.st=fr;tmp.nd.nd=to;
if(tree[i].sum&(1<<(fr-1)))
{
tree[i].sum-=(1<<(fr-1));
tree[i].sum=(tree[i].sum|(1<<(to-1)));
tree[i].to_do.push_back(tmp);
}
return;
}
push_down(i);
int mid=cut(tree[i].l,tree[i].r);
if(mid>=l) change2(lson(i),l,r,fr,to);
if(mid<r) change2(rson(i),l,r,fr,to);
push_up(i);
}
inline int getans(int i,int l,int r)
{
if(tree[i].l>=l&&tree[i].r<=r)
return tree[i].sum;
push_down(i);
int mid=cut(tree[i].l,tree[i].r);short res=0;
if(mid>=l) res=(res|getans(lson(i),l,r));
if(mid<r) res=(res|getans(rson(i),l,r));
return res;
}
int main()
{
n=rd();m=rd();
build(1,1,n);
for(reint i=1;i<=m;i++)
{
id=rd();
if(id==1)
{
a=rd();b=rd();c=rd();
change1(1,a,b,c);
}
else if(id==2)
{
a=rd();b=rd();c=rd();d=rd();
change2(1,a,b,c,d);
}
else
{
a=rd();b=rd();
int ans=getans(1,a,b);
for(short j=1;j<=7;j++)
if(ans&(1<<(j-1))) push(j+'0'),pk;
pn;
}
}
fwrite(pbuf,1,pp-pbuf,stdout);
return 0;
}C96 transformation wsl!
