Gym - 101908L 樹鏈剖分裸題
阿新 • • 發佈:2018-11-11
VJ的連結:https://cn.vjudge.net/problem/Gym-101908L
題目大意:
一棵\(n\)個點的樹上,查詢\(a\) ~ \(b\),\(c\) ~ \(d\)兩段路徑公共的點的個數。
解題思路:
讀完題就感覺是樹鏈剖分的裸題呀...
把每個點的值初始化為\(0\),對於每次詢問,把\(a\) ~ \(b\)這條路徑的值\(+1\),查詢\(c\) ~ \(d\)的區間和,再把\(a\) ~ \(b\)這條路徑的值\(-1\)改回來。\(c\) ~ \(d\)的區間和就是交點數量。
程式碼:
(就是樹鏈剖分維護點權的板子,改了下main就過了...)
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 100000+10; int w[maxn]; int N, Q; struct { int to,next; }e[maxn<<1]; int head[maxn],edgeNum; void add(int u,int v) { e[edgeNum].next = head[u]; e[edgeNum].to = v; head[u] = edgeNum++; } /*-------------------------樹剖------------------------------*/ int deep[maxn],fa[maxn],siz[maxn],son[maxn]; void dfs1(int u,int pre,int d) { deep[u] = d; fa[u] = pre; siz[u] = 1; son[u] = 0; for(int i=head[u];~i;i=e[i].next) { int v = e[i].to; if(v!=pre) { dfs1(v,u,d+1); siz[u] += siz[v]; if(siz[v]>siz[son[u]]) son[u] = v; } } } int top[maxn],id[maxn],rk[maxn],cnt; int &n = cnt; void dfs2(int u,int t) { top[u] = t; id[u] = ++cnt; rk[cnt] = u; if(!son[u]) return; dfs2(son[u],t); for(int i=head[u];~i;i=e[i].next) { int v = e[i].to; if(v!=son[u]&&v!=fa[u]) dfs2(v,v); } } /*-------------------------樹剖------------------------------*/ /*-------------------------線段樹------------------------------*/ #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 int sum[maxn<<2],lazy[maxn<<2]; void pushup(int rt) { sum[rt] = (sum[rt<<1] + sum[rt<<1|1]); } void build(int l,int r,int rt) { if(l==r) { sum[rt] = w[rk[l]]; return ; } int m = l+r>>1; build(lson); build(rson); pushup(rt); } void pushdown(int rt,int l,int r) { if(lazy[rt]) { lazy[rt<<1] = (lazy[rt<<1] + lazy[rt]); lazy[rt<<1|1] = (lazy[rt<<1|1] + lazy[rt]); sum[rt<<1] += (lazy[rt] * l); sum[rt<<1|1] += (lazy[rt] * r); lazy[rt] = 0; } } void update(int L,int R,int val,int l,int r,int rt) { if(L<=l&&r<=R) { sum[rt] = (sum[rt] + (val) * ((r-l+1))); lazy[rt] += val; return ; } int m = l + r >> 1; pushdown(rt,m-l+1,r-m); if(L<=m) update(L,R,val,lson); if(R>m) update(L,R,val,rson); pushup(rt); } int query(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) return sum[rt]; int m = l + r >> 1,ans = 0; pushdown(rt,m-l+1,r-m); if(L<=m) ans = (ans + query(L,R,lson)); if(R>m) ans = (ans + query(L,R,rson)); return ans; } /*-------------------------線段樹------------------------------*/ /*-----------------------樹剖加線段樹--------------------------*/ void update(int x,int y,int z) { while(top[x]!=top[y]) { if(deep[top[x]]<deep[top[y]]) swap(x,y); update(id[top[x]],id[x],z,1,n,1); x = fa[top[x]]; } if(deep[x]>deep[y]) swap(x,y); update(id[x],id[y],z,1,n,1); } int query(int x,int y) { int ans = 0; while(top[x] != top[y]) { if(deep[top[x]] < deep[top[y]]) swap(x,y); ans = (ans + query(id[top[x]],id[x],1,n,1)); x = fa[top[x]]; } if(deep[x]>deep[y]) swap(x,y); ans = (ans + query(id[x],id[y],1,n,1)); return ans; } /*-----------------------樹剖加線段樹--------------------------*/ void init() { memset(head,-1,4*N+4); cnt = edgeNum = 0; } int u, v, x1, y1, x2, y2; int main() { scanf("%d%d",&N,&Q); init(); for(int i=1;i<N;++i) { w[i] = 0; scanf("%d%d",&u,&v); add(u,v); add(v,u); } dfs1(1,0,0); dfs2(1,1); build(1,n,1); while(Q--) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); update(x1,y1,1); printf("%d\n",query(x2,y2)); update(x1,y1,-1); } return 0; }