375. Query on a treeProblem code: QTREE |
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
题解：
抄板子
#include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include<iostream> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int maxn = 100010; const int inf = 0x3f3f3f3f; const int mod = 1000000007; struct sss { int to,next; }edge[maxn >> 1]; int head[maxn],tot; int top[maxn];//top[v]表示v所在的重链的顶端节点 int fa[maxn]; //父亲节点 int deep[maxn];//深度 int num[maxn];//num[v]表示以v为根的子树的节点数 int p[maxn];//p[v]表示v与其父亲节点的连边在线段树中的位置 int fp[maxn];//和p数组相反 int son[maxn];//重儿子 int pos; void init() { mem(head);tot=1;mem(son); pos=1; } void add(int u,int v) { edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++; } void dfs1(int u,int pre,int d) {///第一遍dfs求出fa,deep,num,son deep[u]=d; fa[u]=pre; num[u]=1; for(int i=head[u];i;i=edge[i].next) { int v=edge[i].to; if(v != pre) { dfs1(v,u,d+1); num[u] += num[v]; if(son[u] == 0 || num[v] > num[son[u]]) son[u]=v; } } } void dfs2(int u,int sp) {/////第二遍dfs求出top和p top[u]=sp; if(son[u] != 0) { p[u] = pos++; fp[p[u]] = u; dfs2(son[u],sp); } else { p[u] = pos++; fp[p[u]] = u; return ; } for(int i=head[u];i;i=edge[i].next) { int v=edge[i].to; if(v != son[u] && v != fa[u]) { dfs2(v,v); } } } ////线段树 struct ss{ int l,r; int M; }tr[maxn >> 2]; void build(int k,int s,int t) { tr[k].l=s;tr[k].r=t; tr[k].M=0; if(s==t) return ; int mid = (s+t) >> 1; build(k<<1,s,mid); build(k<<1|1,mid+1,t); } void pushup(int k) { tr[k].M=max(tr[k<<1].M,tr[k<<1|1].M); } void update(int k,int x,int c) { if(tr[k].l==x&&tr[k].r==x) { tr[k].M=c; return ; } int mid = (tr[k].l+tr[k].r) >> 1; if(x<=mid) update(k<<1,x,c); else update(k<<1|1,x,c); pushup(k); } int ask(int k,int s,int t) { if(tr[k].l==s&&tr[k].r==t) { return tr[k].M; } int mid = (tr[k].l+tr[k].r) >> 1; if(t<=mid) return ask(k<<1,s,t); else if(s>mid) return ask(k<<1|1,s,t); else return max(ask(k<<1,s,mid),ask(k<<1|1,mid+1,t)); } int finds(int u,int v) { int fx = top[u],fy = top[v]; int tmp=0; while(fx!=fy) { if(deep[fx] < deep[fy]) { swap(fx,fy); swap(u,v); } tmp = max(tmp,ask(1,p[fx],p[u])); u = fa[fx]; fx = top[u]; } if(u==v) return tmp; if(deep[u] > deep[v]) swap(u,v); return max(tmp,ask(1,p[son[u]],p[v])); } int main() { int T,e[maxn][3],n; scanf("%d",&T); while(T--) { init(); scanf("%d",&n); for(int i=1;i<n;i++) { scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]); add(e[i][0],e[i][1]);add(e[i][1],e[i][0]); } dfs1(1,0,0);// dfs2(1,1);// build(1,1,pos-1); for(int i=1;i<n;i++) { if(deep[e[i][0]] > deep[e[i][1]]) swap(e[i][0],e[i][1]); update(1,p[e[i][1]],e[i][2]); } char ch[10]; int u,v; while(~scanf("%s",ch)) { if(ch[0]=='D') break; scanf("%d%d",&u,&v); if(ch[0]=='Q') printf("%d ",finds(u,v)); else update(1,p[e[u][1]],v); } } return 0; }