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李超树学习记录

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引子

本文是按照李超老师的线段树ppt学习的学习记录以及一些心得

cf145E

题意

两个操作：
1.翻转\([l,r]\)中的0和1
2.求\([l,r]\)的最长不下降子序列长度

思路

线段树维护00,01,10,11的长度，翻转就是交换(00,11)和(01,10)
答案就是max(00-00,00-01,00-11,01-11,11-11)

代码

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

题意

一棵树，初始每条边都是1，两个操作：
1.将某个边改为0
2.问一个点到根的边权和

思路

法一

线段树维护dfs序，操作1为区间修改，操作2为单点查询
这题10s时限，加了快读跑了9s。。

代码

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;
}

vectorv[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;
}

法二

建欧拉序，进入的权值为1，退出的权值为-1
操作1为修改进入和退出的权值为0
操作2为查询前缀和
树状数组即可
跑了5s

代码

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;
}

vectorv[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

标签：const,lc,记录,int,学习,rc,include,root,李超树
来源： https://www.cnblogs.com/wrjlinkkkkkk/p/11515350.html