• 【Luogu】P3950部落冲突(树链剖分)


      题目链接

      状态奇差无比,sbt都能错一遍。

      不动笔光想没有想到怎么做,画图之后发现一个很明显的性质……

      那就是两个开战的部落,其中一个是另一个的父亲。

      所以在儿子那里加个权值。查询的时候树链剖分查询链上点权和,减去lca的点权(因为lca那如果有点权,代表的是lca和lca的父亲之间的那条边)。

      

    #include<cstdio>
    #include<algorithm>
    #include<cctype>
    #include<cstring>
    #include<cstdlib>
    #define left (rt<<1)
    #define right (rt<<1|1)
    #define mid ((l+r)>>1)
    #define lson l,mid,left
    #define rson mid+1,r,right
    #define maxn 300050
    using namespace std;
    inline long long read(){
        long long num=0,f=1;
        char ch=getchar();
        while(!isdigit(ch)){
            if(ch=='-')    f=-1;
            ch=getchar();
        }
        while(isdigit(ch)){
            num=num*10+ch-'0';
            ch=getchar();
        }
        return num*f;
    }
    
    struct Edge{
        int next,to;
    }edge[maxn*3];
    int head[maxn],num;
    inline void add(int from,int to){
        edge[++num]=(Edge){head[from],to};
        head[from]=num;
    }
    
    int tree[maxn*4];
    int dfn[maxn];
    int deep[maxn];
    int father[maxn];
    int size[maxn];
    int top[maxn];
    int son[maxn];
    int ID,n,m;
    
    void unifnd(int x,int fa){
        deep[x]=deep[fa]+1;    size[x]=1;
        for(int i=head[x];i;i=edge[i].next){
            int to=edge[i].to;
            if(to==fa)    continue;
            father[to]=x;
            unifnd(to,x);
            size[x]+=size[to];
            if(son[x]==0||size[son[x]]<size[to])    son[x]=to;
        }
    }
    
    void unionn(int x,int Top){
        dfn[x]=++ID;    top[x]=Top;
        if(son[x]==0)    return;
        unionn(son[x],Top);
        for(int i=head[x];i;i=edge[i].next){
            int to=edge[i].to;
            if(to==father[x]||to==son[x])    continue;
            unionn(to,to);
        }
    }
    
    inline void pushup(int rt){
        tree[rt]=tree[left]+tree[right];
    }
    
    void update(int o,int num,int l,int r,int rt){
        if(l==r){
            tree[rt]+=num;
            return;
        }
        if(o<=mid)    update(o,num,lson);
        else        update(o,num,rson);
        pushup(rt);
        return;
    }
    
    int query(int from,int to,int l,int r,int rt){
        if(from<=l&&to>=r)    return tree[rt];
        int ans=0;
        if(from<=mid)    ans+=query(from,to,lson);
        if(to>mid)        ans+=query(from,to,rson);
        return ans;
    }
    
    struct War{
        int from,to;
    }q[maxn];int cnt;
    
    inline void adds(int from,int to){
        if(deep[from]<deep[to])    swap(from,to);
        q[++cnt]=(War){from,to};
        update(dfn[from],1,1,n,1);
        return;
    }
    
    inline void del(int rnk){
        int from=q[rnk].from;
        update(dfn[from],-1,1,n,1);
    }
    
    inline int LCA(int from,int to){
        while(top[from]!=top[to]){
            if(deep[top[from]]<deep[top[to]])    swap(from,to);
            from=father[top[from]];
        }
        if(deep[from]>deep[to])    swap(from,to);
        return from;
    }
    
    inline int ask(int from,int to){
        int lca=LCA(from,to),ans=-query(dfn[lca],dfn[lca],1,n,1);
        while(top[from]!=top[to]){
            if(deep[top[from]]<deep[top[to]])    swap(from,to);
            ans+=query(dfn[top[from]],dfn[from],1,n,1);
            if(ans>0)    return 0;
            from=father[top[from]];
        }
        if(deep[from]>deep[to])    swap(from,to);
        ans+=query(dfn[from],dfn[to],1,n,1);
        if(ans>0)    return 0;
        return 1;
    }
    
    int main(){
        n=read(),m=read();
        for(int i=1;i<n;++i){
            int x=read(),y=read();
            add(x,y);
            add(y,x);
        }
        unifnd(1,1);
        unionn(1,1);
        for(int i=1;i<=m;++i){
            char c[10];
            scanf("%s",c);
            if(c[0]=='Q'){
                int x=read(),y=read();
                if(ask(x,y))    printf("Yes
    ");
                else            printf("No
    ");
            }
            else if(c[0]=='C'){
                int x=read(),y=read();
                adds(x,y);
            }
            else{
                int x=read();
                del(x);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cellular-automaton/p/8387305.html
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