No more complete

Primer

This article is a study in accordance with the Lie teacher learning segment tree ppt record and some experience

cf145E

The meaning of problems

Two operations:
1. Turn \ ([l, r] \ ) of 0 and 1
2. demand \ ([l, r] \ ) of the longest sequence length is not reduced sub

Thinking

00,01,10,11 length of segment tree maintenance, flipping is the exchange (00,11) and (01,10)
The answer is max (00-00,00-01,00-11,01-11,11-11 )

Code


#include
   
   
    
    
#include
    
    
     
     
#include
     
     
      
      
#include
      
      
       
       
#include
       
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include 
               #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair 
               
                 PI; typedef pair 
                
                  PLL; const db eps = 1e-6; const int mod = 1e9+7; const int maxn = 1e6+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); int len00[maxn<<2],len01[maxn<<2],len11[maxn<<2],len10[maxn<<2]; int flg[maxn<<2]; int a[maxn]; int n,m; void pushup(int l, int r, int root){ int mid = (l+r)>>1; len00[root]=len00[lc]+len00[rc]; len01[root]=max(len00[lc]+max(len01[rc],len11[rc]),len01[lc]+len11[rc]); len10[root]=max(len11[lc]+max(len10[rc],len00[rc]),len10[lc]+len00[rc]); len11[root]=len11[lc]+len11[rc]; return; } void build(int l, int r, int root){ int mid = (l+r)>>1; if(l==r){ if(a[l])len11[root]=1; else len00[root]=1; return; } build(lson);build(rson); pushup(l,r,root); return; } void pushdown(int l, int r, int root){ int mid = (l+r)>>1; if(flg[root]){ flg[lc]^=1;flg[rc]^=1; flg[root]^=1; swap(len00[lc],len11[lc]);swap(len00[rc],len11[rc]); swap(len01[lc],len10[lc]);swap(len01[rc],len10[rc]); } } void update(int x, int y, int l, int r, int root){ int mid = (l+r)>>1; if(x<=l&&r<=y){ flg[root]^=1; swap(len00[root],len11[root]); swap(len01[root],len10[root]); return; } pushdown(l,r,root); if(x<=mid)update(x,y,lson); if(mid 
                 
                   >1; if(l==r)return; pushdown(l,r,root); gao(lson);gao(rson); return; } int main() { scanf("%d %d", &n, &m); for(int i = 1; i <= n; i++){ scanf("%1d", &a[i]); a[i]=(a[i]==4?0:1); } build(1,n,1); for(int i = 1; i <= m; i++){ char op[66]; scanf("%s",op+1); if(op[1]=='s'){ int l, r; scanf("%d %d", &l, &r); update(l,r,1,n,1); } else{ int ans = 0; ans=max(ans,len11[1]); ans=max(ans,len01[1]); ans=max(ans,len00[1]); printf("%d\n",ans); } } return 0; }

BZOJ1103

The meaning of problems

A tree, each edge is initially 1, two operations:
1. 0 to a side
2. Q a point to the right side and the roots

Thinking

A law

Dfs segment tree maintenance procedures, operating range of 1 to modify, operate as a single point of inquiry 2
this question time 10s, plus a quick read run 9s. .

Code


#include
   
   
    
    
#include
    
    
     
     
#include
     
     
      
      
#include
      
      
       
       
#include
       
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include 
               #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair 
               
                 PI; typedef pair 
                
                  PLL; const db eps = 1e-6; const int mod = 1e9+7; const int maxn = 3e5+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); inline int read(){ int num; char ch; while((ch=getchar())<'0' || ch>'9'); num=ch-'0'; while((ch=getchar())>='0' && ch<='9'){ num=num*10+ch-'0'; } return num; } vector 
                 
                   v[maxn]; int n,q; int S[maxn]; int bg[maxn],ed[maxn]; int tot; int a[maxn]; void dfs(int x, int fa, int cnt){ S[++tot]=x; bg[x]=tot; a[tot]=cnt; for(int i = 0; i < (int)v[x].size(); i++){ int y = v[x][i]; if(y==fa)continue; dfs(y,x,cnt+1); } ed[x]=tot; } int sum[maxn<<2],addv[maxn<<2]; void build(int l, int r, int root){ int mid = (l+r)>>1; if(l==r){ sum[root]=a[l]; return; } build(lson);build(rson); sum[root]=sum[lc]+sum[rc]; return; } void pushdown(int l, int r, int root){ int mid = (l+r)>>1; if(addv[root]!=0){ addv[lc]+=addv[root]; addv[rc]+=addv[root]; sum[lc]+=(mid-l+1)*addv[root]; sum[rc]+=(r-mid)*addv[root]; addv[root]=0; } } void update(int x, int y, int add, int l, int r, int root){ int mid = (l+r)>>1; if(x<=l&&r<=y){ addv[root]+=add; sum[root]+=add*(r-l+1); return; } pushdown(l,r,root); if(x<=mid)update(x,y,add,lson); if(y>mid)update(x,y,add,rson); sum[root]=sum[lc]+sum[rc]; return; } int ask(int l, int r, int root, int x){ int mid = (l+r)>>1; if(l==r)return sum[root]; pushdown(l,r,root); if(x<=mid)return ask(lson,x); else return ask(rson,x); } int main(){ n=read(); tot = 0; for(int i = 1; i < n; i++){ int x, y; x=read();y=read(); v[x].pb(y);v[y].pb(x); } dfs(1,-1,0); build(1,n,1); scanf("%d", &q); for(int i = 1; i < q+n; i++){ char op[5]; int x, y; scanf("%s",op+1); if(op[1]=='W'){ x=read(); //scanf("%d", &x); printf("%d\n",ask(1,n,1,bg[x])); } else{ x=read();y=read(); //scanf("%d %d", &x ,&y); if(x>y)swap(x,y); update(bg[y],ed[y],-1,1,n,1); } } return 0; }

Act II

Euler construction sequence, the value 1 into the weight, the weight value of -1 exit
operations 1 to modify the entry and exit of the weight value of 0
operation to query prefix 2 and
Fenwick tree can
run 5s

Code


#include
   
   
    
    
#include
    
    
     
     
#include
     
     
      
      
#include
      
      
       
       
#include
       
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include 
               #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair 
               
                 PI; typedef pair 
                
                  PLL; const db eps = 1e-6; const int mod = 1e9+7; const int maxn = 8e5+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); inline int read(){ int num; char ch; while((ch=getchar())<'0' || ch>'9'); num=ch-'0'; while((ch=getchar())>='0' && ch<='9'){ num=num*10+ch-'0'; } return num; } int lowbit(int x){ return x&-x; } int n,q; int tot; int tree[maxn<<2]; void add(int x,int C){ //printf("***add %d %d\n",x,C); for(int i=x;i<=tot;i+=lowbit(i)){ tree[i]+=C; } } int sum(int x){ int ans=0; //printf("***sum %d\n",x); for(int i=x;i;i-=lowbit(i)){ ans+=tree[i]; } return ans; } vector 
                 
                   v[maxn]; int S[maxn]; int bg[maxn],ed[maxn]; int a[maxn]; void dfs(int x, int fa, int cnt){ S[++tot]=x; bg[x]=tot; a[tot]=1; for(int i = 0; i < (int)v[x].size(); i++){ int y = v[x][i]; if(y==fa)continue; dfs(y,x,cnt+1); } S[++tot]=x; a[tot]=-1; ed[x]=tot; } int main(){ n=read(); tot = 0; for(int i = 1; i < n; i++){ int x, y; x=read();y=read(); v[x].pb(y);v[y].pb(x); } dfs(1,-1,0); for(int i = 1; i <= n; i++){ add(bg[i],1);add(ed[i],-1); } scanf("%d", &q); for(int i = 1; i < q+n; i++){ char op[5]; int x, y; scanf("%s",op+1); if(op[1]=='W'){ x=read(); printf("%d\n",sum(bg[x])-1); } else{ x=read();y=read(); if(x>y)swap(x,y); add(bg[y],-1);add(ed[y],1); } } return 0; }

Code

Li Chao tree learning record

Primer

cf145E

The meaning of problems

Thinking

Code

BZOJ1103

The meaning of problems

Thinking

A law

Code

Act II

Code

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