CF165D Beard Graph

题目链接

树剖题不用多说，一开始所有黑边的权值是-1，若有修改白边的操作，就把白边的值赋为100000。

之后查询边权之和时，如果和大于1000000，就肯定存在白边，直接输出-1。

//做法：树剖，一开始黑边边权全设为1，若有修改白边，设为100000
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
struct node{
    int nxt,to,val,from;
}edge[maxn*4];
int head[maxn],cnt;
void add(int x,int y,int v){
    edge[++cnt].nxt=head[x];
    edge[cnt].from=x;
    edge[cnt].to=y;
    edge[cnt].val=v;
    head[x]=cnt;
}
int n,m;
int x,y;
int opt,num;
int dep[maxn],fa[maxn],size[maxn],son[maxn],w[maxn];
void dfs1(int x,int f){
    fa[x]=f;
    size[x]=1;
    dep[x]=dep[f]+1;
    int maxson=-1;
    for(int i=head[x];i;i=edge[i].nxt){
        int v=edge[i].to;
        if(v==fa[x]) continue;
        w[v]=edge[i].val;
        dfs1(v,x);
        size[x]+=size[v];
        if(size[v]>maxson){
            maxson=size[v];
            son[x]=v;
        }
    }
}
int id[maxn],top[maxn],va[maxn],Time;
void dfs2(int x,int topf){
    top[x]=topf;
    id[x]=++Time;
    va[id[x]]=w[x];
    if(!son[x]) return;
    dfs2(son[x],topf);
    for(int i=head[x];i;i=edge[i].nxt){
        int v=edge[i].to;
        if(v==fa[x]||v==son[x]) continue;
        dfs2(v,v);
    }
}
struct nod{
    int l,r;
    long long sum;
    int lazy;
}tree[maxn*4];
void build(int now,int l,int r){
    tree[now].l=l,tree[now].r=r,tree[now].lazy=-1;
    if(l==r){
        tree[now].sum=va[l];
        return;
    }
    int mid=(l+r)>>1;
    build(now<<1,l,mid);
    build(now<<1|1,mid+1,r);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
void pushdown(int now){
    if(tree[now].lazy!=-1){
        tree[now<<1].sum=(tree[now<<1].r-tree[now<<1].l+1)*tree[now].lazy;
        tree[now<<1|1].sum=(tree[now<<1|1].r-tree[now<<1|1].r+1)*tree[now].lazy;
        tree[now<<1].lazy=tree[now].lazy;
        tree[now<<1|1].lazy=tree[now].lazy;
        tree[now].lazy=-1;
    }
}
void update(int now,int l,int r,int v){
    if(tree[now].l>=l&&tree[now].r<=r){
        tree[now].sum=(tree[now].r-tree[now].l+1)*v;
        tree[now].lazy=v;
        return;
    } 
    pushdown(now);
    int mid=(tree[now].l+tree[now].r)>>1;
    if(l<=mid) update(now<<1,l,r,v);
    if(r>mid) update(now<<1|1,l,r,v);
    tree[now].sum=tree[now<<1].sum+tree[now<<1|1].sum;
}
long long query(int now,int l,int r){
    if(tree[now].l>=l&&tree[now].r<=r) return tree[now].sum;
    pushdown(now);
    int mid=(tree[now].l+tree[now].r)>>1;
    long long val=0;
    if(l<=mid) val+=query(now<<1,l,r);
    if(r>mid) val+=query(now<<1|1,l,r);
    return val;
}
long long link(int x,int y){
    long long ans=0;
    while(top[x]!=top[y]){
        if(dep[top[x]]<dep[top[y]]) swap(x,y);
        ans+=query(1,id[top[x]],id[x]);
        x=fa[top[x]];
    }
    if(dep[x]<dep[y]) swap(x,y);
    ans+=query(1,id[y]+1,id[x]);
    return ans;
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        scanf("%d%d",&x,&y);
        add(x,y,1);add(y,x,1);
    }
    scanf("%d",&m);
    dfs1(1,0);
    dfs2(1,1);
    build(1,1,n);
    for(int i=1;i<=m;i++){
        scanf("%d",&opt);
        if(opt==2){
            scanf("%d",&num);
            int s=edge[num*2].from;
            int t=edge[num*2].to;
            if(dep[s]>dep[t]) swap(s,t);
            update(1,id[t],id[t],1000000);
        } 
        else if(opt==1){
            scanf("%d",&num);
            int s=edge[num*2].from;
            int t=edge[num*2].to;
            if(dep[s]>dep[t]) swap(s,t);
            update(1,id[t],id[t],1);
        }
        else{
            scanf("%d%d",&x,&y);
            if(link(x,y)>=1000000) printf("-1\n");
            else printf("%lld\n",link(x,y));
        }
    }
    return 0;
}