3553: [Shoi2014]三叉神经树(树链剖分)

这道题特别恶心,首先我们可以发现更改的就是出现连续的一或二,那么就用线段树+树链剖分找到这个范围

想到是不难想,就是打起来恶心罢了= =

CODE:

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

#include<stack>

using namespace std;

#define maxn 500100 

struct edges{

int to,next;

}edge[maxn*3];

int l,next[maxn];

int addedge(int x,int y) {

edge[++l]=(edges){y,next[x]};next[x]=l;

}

int s[maxn],ch[maxn],fa[maxn*3],ans[maxn*3],q[maxn];

int bfs() {

int r=0;

q[++r]=1;

int l=0;

while (l<r) {

int u=q[++l];s[u]=1;

for (int i=next[u];i;i=edge[i].next) {

fa[edge[i].to]=u;q[++r]=edge[i].to;

}

}

for (int i=r;i;i--) {

s[fa[q[i]]]+=s[q[i]];

ch[fa[q[i]]]=s[ch[fa[q[i]]]]<s[q[i]]?q[i]:ch[fa[q[i]]];

ans[fa[q[i]]]+=ans[q[i]]>1?1:0;

}

}

int add[maxn],pos[maxn],pre[maxn];

int cnt;

bool b[maxn];

int heavy(int x,bool y) {

stack<int > s;

s.push(1);

add[pos[++cnt]=1]=cnt;

pre[1]=1;

while (!s.empty()) {

int x=s.top();s.pop();

if (!b[x]&&ch[x]) {

s.push(x);

s.push(ch[x]);

add[pos[++cnt]=ch[x]]=cnt;

pre[ch[x]]=pre[x];

b[x]=1;

continue;

}

for (int i=next[x];i;i=edge[i].next) {

if (edge[i].to!=ch[x]) {

s.push(x);

s.push(edge[i].to);

add[pos[++cnt]=edge[i].to]=cnt;

pre[edge[i].to]=edge[i].to;

next[x]=edge[i].next;

break;

}

}

}

return 0;

}

struct bo {

int sz,s,len;

};

struct node{

int l,r,lz;bo b;

}t[maxn*8];

#define s(x) t[x].b.s

#define sz(x) t[x].b.sz

#define len(x) t[x].b.len

bo update(bo l,bo r) {

bo ans;

ans.sz=l.sz+r.sz;

ans.s=r.s;

ans.len=r.len;

if (l.s==r.s&&r.len==r.sz) ans.len+=l.len;

return ans;

}

#define lc(x) (x<<1)

#define rc(x) ((x<<1)+1)

#define mid ((r+l)>>1)

#define upd(x) t[x].b=update(t[lc(x)].b,t[rc(x)].b)

void build(int x,int l,int r) {

t[x].l=l;t[x].r=r;

sz(x)=r-l+1;

if (l==r) {s(x)=ans[pos[l]];len(x)=1;return ;}

build(lc(x),l,mid);build(rc(x),mid+1,r);

upd(x);

}

#define lz(x) t[x].lz

void pushback(int x) {

if (t[x].lz==0) return;

int l=lc(x),r=rc(x);

lz(l)+=lz(x);lz(r)+=lz(x);

s(l)+=lz(x);s(r)+=lz(x);

lz(x)=0;

}

bo que(int x,int y) {

int l=t[x].l,r=t[x].r;

if (r<=y) return t[x].b;

pushback(x);

if (y<=mid) return que(lc(x),y);

if (y>mid) return update(que(lc(x),y),que(rc(x),y));

}

void inc(int x,int x1,int y1,int z) {

int l=t[x].l,r=t[x].r;

if (y1<l||x1>r) return ;

if (x1<=l&&r<=y1) {

t[x].lz+=z;

s(x)+=z;

return ;

}

pushback(x);

inc(lc(x),x1,y1,z);inc(rc(x),x1,y1,z);

upd(x);

}

void gen(int x,int flag) {

while (x) {

bo tmp=que(1,add[x]);

switch (flag) {

case 0:

if (tmp.s==2) {

if (tmp.len>=add[x]-add[pre[x]]+1) {

inc(1,add[pre[x]],add[x],-1);

x=fa[pre[x]];

} else {

inc(1,add[x]-tmp.len,add[x],-1);

x=0;

}

}else {

inc(1,add[x],add[x],-1);

x=0;

}

break;

case 1:

if (tmp.s==1) {

if (tmp.len>=add[x]-add[pre[x]]+1) {

inc(1,add[pre[x]],add[x],1);

x=fa[pre[x]];

} else {

inc(1,add[x]-tmp.len,add[x],1);

x=0;

}

}else {

inc(1,add[x],add[x],1);

x=0;

}

break;

}

}

return ;

}

int main(){

int n;

scanf("%d",&n);

for (int i=1;i<=n;i++) {

int x1,x2,x3;

scanf("%d%d%d",&x1,&x2,&x3);

if (x1<=n) addedge(i,x1);else fa[x1]=i;

if (x2<=n) addedge(i,x2);else fa[x2]=i;

if (x3<=n) addedge(i,x3);else fa[x3]=i;

}

for (int i=1;i<=2*n+1;i++) {

scanf("%d",&ans[i+n]);

ans[fa[i+n]]+=ans[i+n];

}bfs();

heavy(1,1);

build(1,1,n);

int Q;

scanf("%d",&Q);  

while (Q--) {

int x;

scanf("%d",&x);

ans[x]^=1;

gen(fa[x],ans[x]);

printf("%d ",que(1,1).s>=2?1:0);

}

return 0;

}


原文地址:https://www.cnblogs.com/New-Godess/p/4348894.html