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spoj 375 Query on a tree (树链剖分)
Query on a tree
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:
- CHANGE i ti : change the cost of the i-th edge to ti
or - QUERY a b : ask for the maximum edge cost on the path from node a to node b
Input
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
- In the first line there is an integer N (N <= 10000),
- In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
- The next lines contain instructions "CHANGE i ti" or "QUERY a b",
- The end of each test case is signified by the string "DONE".
There is one blank line between successive tests.
Output
For each "QUERY" operation, write one integer representing its result.
Example
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)
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