首页 > 代码库 > AC日记——[USACO11DEC]牧草种植Grass Planting 洛谷 P3038
AC日记——[USACO11DEC]牧草种植Grass Planting 洛谷 P3038
题目描述
Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on the roads between pastures. Farmer John loves Bessie very much, and today he is finally going to plant grass on the roads. He will do so using a procedure consisting of M steps (1 <= M <= 100,000).
At each step one of two things will happen:
FJ will choose two pastures, and plant a patch of grass along each road in between the two pastures, or,
- Bessie will ask about how many patches of grass on a particular road, and Farmer John must answer her question.
Farmer John is a very poor counter -- help him answer Bessie‘s questions!
给出一棵n个节点的树，有m个操作，操作为将一条路径上的边权加一或询问某条边的权值。
输入输出格式
输入格式：
Line 1: Two space-separated integers N and M
Lines 2..N: Two space-separated integers describing the endpoints of a road.
- Lines N+1..N+M: Line i+1 describes step i. The first character of the line is either P or Q, which describes whether or not FJ is planting grass or simply querying. This is followed by two space-separated integers A_i and B_i (1 <= A_i, B_i <= N) which describe FJ‘s action or query.
输出格式：
- Lines 1..???: Each line has the answer to a query, appearing in the same order as the queries appear in the input.
输入输出样例
输入样例#1：
4 6 1 4 2 4 3 4 P 2 3 P 1 3 Q 3 4 P 1 4 Q 2 4 Q 1 4
输出样例#1：
2 1 2
思路：
裸树剖；
来，上代码：
#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define maxn 100005using namespace std;struct TreeNodeType { int l,r,dis,mid,flag;};struct TreeNodeType tree[maxn<<2];struct EdgeType { int to,next;};struct EdgeType edge[maxn<<1];int if_z,n,m,cnt,deep[maxn],f[maxn],size[maxn];int top[maxn],id[maxn],head[maxn];char Cget;inline void in(int &now){ now=0,if_z=1,Cget=getchar(); while(Cget>‘9‘||Cget<‘0‘) { if(Cget==‘-‘) if_z=-1; Cget=getchar(); } while(Cget>=‘0‘&&Cget<=‘9‘) { now=now*10+Cget-‘0‘; Cget=getchar(); } now*=if_z;}void search_1(int now,int fa){ int pos=cnt++; deep[now]=deep[fa]+1,f[now]=fa; for(int i=head[now];i;i=edge[i].next) { if(edge[i].to==fa) continue; search_1(edge[i].to,now); } size[now]=cnt-pos;}void search_2(int now,int chain){ int pos=0; top[now]=chain,id[now]=++cnt; for(int i=head[now];i;i=edge[i].next) { if(edge[i].to==f[now]) continue; if(size[edge[i].to]>size[pos]) pos=edge[i].to; } if(pos==0) return ; search_2(pos,chain); for(int i=head[now];i;i=edge[i].next) { if(edge[i].to==f[now]||edge[i].to==pos) continue; search_2(edge[i].to,edge[i].to); }}void tree_build(int now,int l,int r){ tree[now].l=l,tree[now].r=r; if(l==r) return ; tree[now].mid=(l+r)>>1; tree_build(now<<1,l,tree[now].mid); tree_build(now<<1|1,tree[now].mid+1,r);}void tree_change(int now,int l,int r){ if(tree[now].l==l&&tree[now].r==r) { tree[now].dis+=r-l+1; tree[now].flag++; return ; } if(tree[now].flag) { tree[now<<1].dis+=tree[now].flag*(tree[now].mid-tree[now].l+1); tree[now<<1|1].dis+=tree[now].flag*(tree[now].r-tree[now].mid); tree[now<<1].flag+=tree[now].flag,tree[now<<1|1].flag+=tree[now].flag; tree[now].flag=0; } if(l>tree[now].mid) tree_change(now<<1|1,l,r); else if(r<=tree[now].mid) tree_change(now<<1,l,r); else { tree_change(now<<1,l,tree[now].mid); tree_change(now<<1|1,tree[now].mid+1,r); } tree[now].dis=tree[now<<1].dis+tree[now<<1|1].dis;}int tree_query(int now,int l,int r){ if(tree[now].l==l&&tree[now].r==r) return tree[now].dis; if(tree[now].flag) { tree[now<<1].dis+=tree[now].flag*(tree[now].mid-tree[now].l+1); tree[now<<1|1].dis+=tree[now].flag*(tree[now].r-tree[now].mid); tree[now<<1].flag+=tree[now].flag,tree[now<<1|1].flag+=tree[now].flag; tree[now].flag=0; } if(l>tree[now].mid) return tree_query(now<<1|1,l,r); else if(r<=tree[now].mid) return tree_query(now<<1,l,r); else return tree_query(now<<1,l,tree[now].mid)+tree_query(now<<1|1,tree[now].mid+1,r);}int main(){ in(n),in(m);int u,v; for(int i=1;i<n;i++) { in(u),in(v); edge[++cnt].to=v,edge[cnt].next=head[u],head[u]=cnt; edge[++cnt].to=u,edge[cnt].next=head[v],head[v]=cnt; } char type; cnt=0,search_1(1,0); cnt=0,search_2(1,1); tree_build(1,1,n); while(m--) { cin>>type;in(u),in(v); if(type==‘P‘) { while(top[u]!=top[v]) { if(deep[top[u]]<deep[top[v]]) swap(u,v); tree_change(1,id[top[u]],id[u]); u=f[top[u]]; } if(u==v) continue; if(deep[u]>deep[v]) swap(u,v); tree_change(1,id[u]+1,id[v]); } else { int pos=0; while(top[u]!=top[v]) { if(deep[top[u]]<deep[top[v]]) swap(u,v); pos+=tree_query(1,id[top[u]],id[u]); u=f[top[u]]; } if(u==v) { printf("%d\n",pos); continue; } if(deep[u]>deep[v]) swap(u,v); printf("%d\n",pos+tree_query(1,id[u]+1,id[v])); } } return 0;}
AC日记——[USACO11DEC]牧草种植Grass Planting 洛谷 P3038
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