luogu P3950 部落冲突

版权声明：thanks for your copy! https://blog.csdn.net/weixin_43907802/article/details/89555455

analysis

分析可知，用树剖

关键在于将边权转化为点权

我们可以将边权向下压，压到两边的点中深度较大的的那个点，然后通过求路径上的点权值和是否为0来判断是否可通过（注意要权值和要减去LCA的权值，因为路径并不会经过LCA和它上面那个点之间的连边）

真的真的不要把线段树和树剖打爆了啊啊啊啊啊啊啊啊啊

code

#include<bits/stdc++.h>
using namespace std;
#define loop(i,start,end) for(register int i=start;i<=end;++i)
#define anti_loop(i,start,end) for(register int i=start;i>=end;--i)
#define clean(arry,num) memset(arry,num,sizeof(arry))
#define max(a,b) ((a>b)?a:b)
#define min(a,b) ((a<b)?a:b)
#define ll long long
const int maxn=300000+10,maxm=300000+10;
int n,m,cnt=0;
template<typename T>void read(T &x){
    x=0;char r=getchar();T neg=1;
    while(r>'9'||r<'0'){if(r=='-')neg=-1;r=getchar();}
    while(r>='0'&&r<='9'){x=(x<<1)+(x<<3)+r-'0';r=getchar();}
    x*=neg;
}
int dep[maxn],son[maxn],fa[maxn],siz[maxn];
int seg[maxn],rev[maxn],top[maxn];
struct node{
    int e;int nxt;
}edge[maxn<<1];
int head[maxn];
int sum[maxn<<2],lazy[maxn<<2];
struct oppo{int a;int b;};
int warx[maxn],wary[maxn];
inline void addl(int u,int v){
   edge[cnt].e=v;
   edge[cnt].nxt=head[u];
   head[u]=cnt++;
}
void dfs1(int u,int f){
        dep[u]=dep[f]+1;
        siz[u]=1;fa[u]=f;
        son[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].e;
            if(v==f)continue;
            dfs1(v,u);
            siz[u]+=siz[v];
            if(siz[v]>siz[son[u]])son[u]=v;
        }
}
void dfs2(int u,int f){
    if(son[u]){
        seg[son[u]]=++seg[0];
        rev[seg[0]]=son[u];
        top[son[u]]=top[u];
        dfs2(son[u],u);
    }
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].e;
        if(!top[v]){
            seg[v]=++seg[0];
            rev[seg[0]]=v;
            top[v]=v;
            dfs2(v,u);
        }
    }
}
inline void buildtree(){clean(sum,0);}
void update(int p,int nl,int nr,int rt,int w){
    if(nl==nr&&nr==p){
        sum[rt]+=w;
        return;
    }
    int mid=((nr+nl)>>1);
    if(mid>=p)update(p,nl,mid,rt<<1,w);
    else update(p,mid+1,nr,rt<<1|1,w);
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
int query(int l,int r,int nl,int nr,int rt){
    if(l<=nl&&nr<=r){
        return sum[rt];
    }
    int mid=((nr+nl)>>1);
    int _sum=0;
    if(mid>=l)_sum+=query(l,r,nl,mid,rt<<1);
    if(mid<r)_sum+=query(l,r,mid+1,nr,rt<<1|1);
    return _sum;
}
inline void reply(){
    int numofWar=0;
    loop(i,1,m){
        register char op[100];
        scanf("%s",op);
        if(op[0]=='Q'){
            int x,y;read(x),read(y);
            int fx=top[x],fy=top[y];
            //bool access=true;
            int _sum=0;
            while(fx!=fy){
                if(dep[fx]<dep[fy])swap(fx,fy),swap(x,y);
                _sum+=query(seg[fx],seg[x],1,seg[0],1);
                x=fa[fx],fx=top[x];
            }
            if(dep[x]<dep[y])swap(x,y);
            _sum+=query(seg[y],seg[x],1,seg[0],1);
            _sum-=query(seg[y],seg[y],1,seg[0],1);
            if(_sum>0)printf("No\n");
            else printf("Yes\n");
        }
        else if(op[0]=='C'){
            int x,y;read(x),read(y);
            warx[++numofWar]=x;
            wary[numofWar]=y;
            update(seg[((dep[x]<dep[y])?y:x)],1,seg[0],1,1);//update(((dep[x]<dep[y])?y:x),1,seg[0],1,1);
        }
        else if(op[0]=='U'){
            int x;read(x);
            update(seg[((dep[wary[x]]>dep[warx[x]])?wary[x]:warx[x])],1,seg[0],1,-1);//update(((dep[wary[x]]>dep[warx[x]])?wary[x]:warx[x]),1,seg[0],1,-1);
        }
    }
}
int main(){
    #ifndef ONLINE_JUDGE
    freopen("datain2.txt","r",stdin);
    #endif
    clean(dep,0),clean(head,-1),clean(son,0),clean(fa,0),clean(siz,0),clean(seg,0),clean(rev,0),clean(top,0);
    read(n),read(m);
    loop(i,1,n-1){
        int xi,yi;read(xi),read(yi);
        addl(xi,yi),addl(yi,xi);
    }
    dfs1(1,0);
    seg[1]=seg[0]=rev[1]=top[1]=1;
    dfs2(1,0);
    //loop(i,1,n)printf("point:%d dep:%d top:%d son:%d siz:%d seg:%d rev:%d fa:%d\n",i,dep[i],top[i],son[i],siz[i],seg[i],rev[i],fa[i]);
    reply();
    return 0;
}