BZOJ 3589 动态树

题意：给一棵树，点有点权，要求支持

1.子树点权+

2.询问k(k<=5)条链并集点权和

题目中所给链全部为某点到根路径上的一段

Sol：

子树点权加？DFS序

再来个询问链上点权和？树链剖分+DFS序

k条并集点权和？观察到k比较小，果断容斥

由数学必修一知识可知，若干集合交集=每个集合-每两个集合并+每三个集合并...

ans=Sum{一条}-Sum{两条交}+Sum{三链交}...

复杂度O(m*log^2(n)*2^k)

一如既往的乱七八糟，字面意义上的

题目要求答案mod(2^31)，我们可以用int自然溢出最后答案&(2147483647)  (就是2^31 -1 啦

Code：

#include<bits/stdc++.h>#define debug(x) cout<<#x<<"="<<x<<endl#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int,int> List;const int maxn = 200009;const int uuz = 2147483647;int n,m;int first[maxn];struct edg{int next;int to;}e[maxn<<1];int e_sum;int fa[maxn],dep[maxn],son[maxn],siz[maxn],pos[maxn],top[maxn],L[maxn],R[maxn];int cnt,ans;struct sg_tree{int sum,tag;}node[maxn<<2];List chain[9];inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}inline void add_edg(int x,int y){e_sum++;e[e_sum].next=first[x];first[x]=e_sum;e[e_sum].to=y;}void dfs1(int x,int f){dep[x]=dep[f]+1;siz[x]=1;fa[x]=f;for(int i=first[x];i;i=e[i].next){int w=e[i].to;if(w==f) continue;dfs1(w,x);siz[x]+=siz[w];if(siz[w]>siz[son[x]]) son[x]=w;}}void dfs2(int x,int t){top[x]=t;pos[x]=++cnt;L[x]=cnt;if(son[x]) dfs2(son[x],t);for(int i=first[x];i;i=e[i].next){int w=e[i].to;if(w==fa[x]||w==son[x]) continue;dfs2(w,w);}R[x]=cnt;}void updata(int rt){node[rt].sum=node[rt<<1].sum+node[rt<<1|1].sum;}void make_plus(int l,int r,int rt,int delta){node[rt].sum+=(r-l+1)*delta;node[rt].tag+=delta;}void down(int l,int r,int rt){if(node[rt].tag){int mid=(l+r)>>1;make_plus(lson,node[rt].tag);make_plus(rson,node[rt].tag);node[rt].tag=0;}}void modify(int l,int r,int rt,int left,int right,int delta){if(l==left&&right==r){make_plus(l,r,rt,delta);return ;}int mid=(l+r)>>1;down(l,r,rt);if(right<=mid) modify(lson,left,right,delta);else if(left>mid) modify(rson,left,right,delta);else modify(lson,left,mid,delta),modify(rson,mid+1,right,delta);updata(rt);}int query(int l,int r,int rt,int left,int right){if(left<=l&&r<=right) return node[rt].sum;int mid=(l+r)>>1;down(l,r,rt);if(right<=mid) return query(lson,left,right);else if(left>mid) return query(rson,left,right);else return query(lson,left,mid)+query(rson,mid+1,right);}int lca(int x,int y){while(top[x]!=top[y]){if(dep[top[x]]<dep[top[y]]) swap(x,y);x=fa[top[x]];}if(dep[x]<dep[y]) return x;return y;}int ask(int x,int y){if(x<=0) return 0;int sum=0;while(top[x]!=top[y]){if(dep[top[x]]<dep[top[y]]) swap(x,y);sum+=query(1,n,1,pos[top[x]],pos[x]);x=fa[top[x]];}if(dep[x]<dep[y]) swap(x,y);sum+=query(1,n,1,pos[y],pos[x]);return sum;}List calc(List x,List y){if(x.first==0) return y;if(x.first==-1) return x;int anc=lca(x.second,y.second);if(dep[anc]<max(dep[x.first],dep[y.first])) return make_pair(-1,-1);if(dep[x.first]>dep[y.first]) return make_pair(x.first,anc);else return make_pair(y.first,anc);}void solve(int step,List now,int opt,int lim){if(step==lim){ans+=ask(now.first,now.second)*opt;return ;}solve(step+1,now,opt,lim);solve(step+1,calc(now,chain[step+1]),-opt,lim);}int main(){n=read();for(int i=1;i<n;i++){int x=read(),y=read();add_edg(x,y);add_edg(y,x);}dfs1(1,0);dfs2(1,1);m=read();while(m--){int opt=read();if(opt==0){int x=read(),k=read();modify(1,n,1,L[x],R[x],k);}else{int k=read();ans=0;for(int i=1;i<=k;i++){int x=read(),y=read();if(dep[x]>dep[y]) swap(x,y);chain[i]=make_pair(x,y);}solve(0,make_pair(0,0),-1,k);printf("%d\n",ans&uuz);}}return 0;}