标签：name ios max pre struct namespace sizeof detail continue
You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.
We will ask you to perfrom some instructions of the following form:
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
There is one blank line between successive tests.
For each "QUERY" operation, write one integer representing its result.
Input: 1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE Output: 1 3
第一次接触树链剖分，貌似是用来处理对树的边权的多次询问，然后对边权进行编号，转化为节点之间的询问。具体关于树链剖分的解析见 http://blog.csdn.net/acdreamers/article/details/10591443
#include <iostream> #include <cstring> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #define met(a,b) memset(a,b,sizeof a) #define pb push_back #define lson(x) ((x<<1)) #define rson(x) ((x<<1)+1) using namespace std; typedef long long ll; const int N=1e5+50; const int M=N*N+10; int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N]; //top 最近的重链父节点 int num; vector<int> v[N]; struct tree { int x,y,val; void read() { scanf("%d%d%d",&x,&y,&val); } }; tree e[N]; void dfs1(int u, int f, int d) { dep[u] = d; siz[u] = 1; son[u] = 0; fa[u] = f; for (int i = 0; i < v[u].size(); i++) { int ff = v[u][i]; if (ff == f) continue; dfs1(ff, u, d + 1); siz[u] += siz[ff]; if (siz[son[u]] < siz[ff]) son[u] = ff; } } void dfs2(int u, int tp) { top[u] = tp; id[u] = ++num; if (son[u]) dfs2(son[u], tp); for (int i = 0; i < v[u].size(); i++) { int ff = v[u][i]; if (ff == fa[u] || ff == son[u]) continue; dfs2(ff, ff); } } struct Tree { int l,r,val; }; Tree tree[4*N]; void pushup(int x) { tree[x].val = max(tree[lson(x)].val, tree[rson(x)].val); } void build(int l,int r,int v) { tree[v].l=l; tree[v].r=r; if(l==r) { tree[v].val = val[l]; return ; } int mid=(l+r)>>1; build(l,mid,v*2); build(mid+1,r,v*2+1); pushup(v); } void update(int o,int v,int val) { //log(n) if(tree[o].l==tree[o].r) { tree[o].val = val; return ; } int mid = (tree[o].l+tree[o].r)/2; if(v<=mid) update(o*2,v,val); else update(o*2+1,v,val); pushup(o); } int query(int x,int l, int r) { if (tree[x].l >= l && tree[x].r <= r) { return tree[x].val; } int mid = (tree[x].l + tree[x].r) / 2; int ans = 0; if (l <= mid) ans = max(ans, query(lson(x),l,r)); if (r > mid) ans = max(ans, query(rson(x),l,r)); return ans; } int Yougth(int u, int v) { int tp1 = top[u], tp2 = top[v]; int ans = 0; while (tp1 != tp2) { if (dep[tp1] < dep[tp2]) { swap(tp1, tp2); swap(u, v); } ans = max(query(1,id[tp1], id[u]), ans); u = fa[tp1]; tp1 = top[u]; } if (u == v) return ans; if (dep[u] > dep[v]) swap(u, v); ans = max(query(1,id[son[u]], id[v]), ans); return ans; } void Clear(int n) { for(int i=1; i<=n; i++) v[i].clear(); } int main() { int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); for(int i=1; i<n; i++) { e[i].read(); v[e[i].x].push_back(e[i].y); v[e[i].y].push_back(e[i].x); } num = 0; dfs1(1,0,1); dfs2(1,1); for (int i = 1; i < n; i++) { if (dep[e[i].x] < dep[e[i].y]) swap(e[i].x, e[i].y); val[id[e[i].x]] = e[i].val; } build(1,num,1); char s[200]; while(~scanf("%s",&s) && s[0]!=‘D‘) { int x,y; scanf("%d%d",&x,&y); if(s[0]==‘Q‘) printf("%d\n",Yougth(x,y)); if (s[0] == ‘C‘) update(1,id[e[x].x],y); } Clear(n); } return 0; }
spoj 375 Query on a tree (树链剖分)
标签：name ios max pre struct namespace sizeof detail continue
原文地址：http://www.cnblogs.com/jianrenfang/p/6354296.html
