题解【[USACO18FEB]New Barns 】

浅谈一下对于这题做完之后的感受（不看题解也是敲不出来啊qwq……）

题意翻译

Farmer John注意到他的奶牛们如果被关得太紧就容易吵架，所以他想开放一些新的牛棚来分散她们。每当FJ建造一个新牛棚的时候，他会将这个牛棚用至多一条双向道路与一个现有的牛棚连接起来。为了确保他的奶牛们足够分散，他有时想要确定从某个特定的牛棚出发，到它能够到达的最远的牛棚的距离（两个牛棚之间的距离等于从一个牛棚出发到另一个之间必须经过的道路条数）。

FJ总共会给出 $\leq Q \leq 10^51≤Q≤105）次请求，每个请求都是“建造”和“距离”之一。对于一个建造请求，FJ建造一个牛棚，并将其与至多一个现有的牛棚连接起来。对于一个距离请求，FJ向你询问从某个特定的牛棚通过一些道路到离它最远的牛棚的距离。保证询问的牛棚已经被建造。请帮助FJ响应所有的请求。$

输入格式（文件名：newbarn.in）：第一行包含一个整数

题目描述

Farmer John notices that his cows tend to get into arguments if they are packed too closely together, so he wants to open a series of new barns to help spread them out. Whenever FJ constructs a new barn, he connects it with at most one bidirectional pathway to an existing barn. In order to make sure his cows are spread sufficiently far apart, he sometimes wants to determine the distance from a certain barn to the farthest possible barn reachable from it (the distance between two barns is the number of paths one must traverse to go from one barn to the other).

FJ will give a total of $\leq Q \leq 10^51≤Q≤105) queries, each either of the form "build" or "distance". For a build query, FJ builds a barn and links it with at most one previously built barn. For a distance query, FJ asks you the distance from a certain barn to the farthest barn reachable from it via a series of pathways. It is guaranteed that the queried barn has already been built. Please help FJ answer all of these queries.$

输入输出格式

输入格式：

The first line contains the integer

输出格式：

Please write one line of output for each distance query. Note that a barn which is connected to no other barns has farthest distance

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,m,opt,a,b,dis[500000];//dis 保存距离 
int bel[500000];//并查集 
struct node{
    int fa,ch[2],sum,tag;
}tr[400000];
struct Node{
    int a,b;
}s[500000];//本题要求合并一个点和一个联通块，连着俩点保存一下 
char cg[3];
inline bool root(int x){
    int g=tr[x].fa;
    return !(tr[g].ch[0]==x||tr[g].ch[1]==x);
}inline void pushup(int x){tr[x].sum=1+tr[tr[x].ch[0]].sum+tr[tr[x].ch[1]].sum;}
inline void pushr(int x){
    if(!x)return;
    swap(tr[x].ch[0],tr[x].ch[1]);
    tr[x].tag^=1;
}inline void pushdown(int x){
    if(tr[x].tag){
        pushr(tr[x].ch[0]);
        pushr(tr[x].ch[1]);
        tr[x].tag=0;
    }
}inline void push(int x){
    if(!root(x))push(tr[x].fa);
    pushdown(x);
}inline void rotate(int x){
    int y=tr[x].fa,z=tr[y].fa,k=tr[y].ch[1]==x;
    if(!root(y))tr[z].ch[tr[z].ch[1]==y]=x;
    tr[x].fa=z;tr[y].ch[k]=tr[x].ch[k^1];
    if(tr[x].ch[k^1])tr[tr[x].ch[k^1]].fa=y;
    tr[y].fa=x;tr[x].ch[k^1]=y;
    pushup(y);pushup(x);
}inline void splay(int x){
    int y,z;
    push(x);
    while(!root(x)){
        y=tr[x].fa,z=tr[y].fa;
        if(!root(y))(tr[z].ch[0]==y)^(tr[y].ch[0]==x)?rotate(x):rotate(y);
        rotate(x);
    }pushup(x);
}inline void access(int x){for(int y=0;x;y=x,x=tr[x].fa){splay(x);tr[x].ch[1]=y;pushup(x);}}
inline void makeroot(int x){
    access(x);splay(x);pushr(x);
}inline void split(int x,int y){
    makeroot(x);
    access(y);splay(y);
}inline void link(int x,int y){
    makeroot(x);tr[x].fa=y;
}inline void cut(int x,int y){
    makeroot(x);access(y);splay(y);
    if(tr[y].ch[0]!=x||tr[x].ch[1])return;
    tr[x].fa=tr[y].ch[0]=0;
    pushup(x);
}inline int find(int x){
    return x==bel[x]?x:bel[x]=find(bel[x]); 
} 
int main(){
    scanf("%d",&n);//其它都是模板qwq…… 
    for(int i=1;i<=n;i++)tr[i].sum=1,bel[i]=i,s[i]={i,i};//初始化 
    for(int i=1;i<=n;++i){
        cin>>cg;
        scanf("%d",&a);
        if(cg[0]=='B'){
            ++opt;
            if(a==-1)continue;
            else{
                int d=find(a);
                link(opt,a);
                bel[opt]=d;//并查集更新 
                split(opt,s[d].a);//每次合并点查询距离 
                int x1=tr[s[d].a].sum-1;
                split(opt,s[d].b);
                int x2=tr[s[d].b].sum-1;
                if(dis[d]<x1)dis[d]=x1,s[d]={opt,s[d].a};//更新每次的最大距离，一起更新s结构体 
                if(dis[d]<x2)dis[d]=x2,s[d]={opt,s[d].b};
            }
        }else{
            int d=find(a);
            split(a,s[d].a);//s是a所连的两个点 
            int x1=tr[s[d].a].sum-1;//求当前距离，一个点连着两个点，求两次取最大 
            split(a,s[d].b);
            int x2=tr[s[d].b].sum-1;
            printf("%d\n",max(x1,x2));//取最大咯 
        }
    }
    return 0;//完结撒花~ 
}

有不当之处也还请看到的各位dalao指出。

6.13地理中考加油！