A - Query on a tree
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87378#problem/A
Description
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.
Sample Input
1
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
Sample Output
1
3
HINT
题意
给一棵树，有两个操作，修改边权，查询u->v中的边的最大值
题解：
树链剖分裸题，但是我还是不会写
代码来自kuangbin
代码：
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> using namespace std; const int MAXN = 10010; struct Edge { int to,next; }edge[MAXN*2]; 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() { tot = 0; memset(head,-1,sizeof(head)); pos = 0; memset(son,-1,sizeof(son)); } void addedge(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 != -1; i = edge[i].next) { int v = edge[i].to; if(v != pre) { dfs1(v,u,d+1); num[u] += num[v]; if(son[u] == -1 || num[v] > num[son[u]]) son[u] = v; } } } void getpos(int u,int sp) //第二遍dfs求出top和p { top[u] = sp; if(son[u] != -1) { p[u] = pos++; fp[p[u]] = u; getpos(son[u],sp); } else { p[u] = pos++; fp[p[u]] = u; return; } for(int i = head[u] ; i != -1; i = edge[i].next) { int v = edge[i].to; if(v != son[u] && v != fa[u]) getpos(v,v); } } //线段树 struct Node { int l,r; int Max; }segTree[MAXN*4]; void build(int i,int l,int r) { segTree[i].l = l; segTree[i].r = r; segTree[i].Max = 0; if(l == r)return; int mid = (l+r)/2; build(i<<1,l,mid); build((i<<1)|1,mid+1,r); } void update(int i,int k,int val) // 更新线段树的第k个值为val { if(segTree[i].l == k && segTree[i].r == k) { segTree[i].Max = val; return; } int mid = (segTree[i].l + segTree[i].r)/2; if(k <= mid)update(i<<1,k,val); else update((i<<1)|1,k,val); segTree[i].Max=max(segTree[i<<1].Max,segTree[i<<1|1].Max); } int query(int i,int l,int r) //查询线段树中[l,r] 的最大值 { if(segTree[i].l == l && segTree[i].r == r) return segTree[i].Max; int mid = (segTree[i].l + segTree[i].r)/2; if(r <= mid)return query(i<<1,l,r); else if(l > mid)return query((i<<1)|1,l,r); else return max(query(i<<1,l,mid),query((i<<1)|1,mid+1,r)); } int find(int u,int v)//查询u->v边的最大值 { int f1 = top[u], f2 = top[v]; int tmp = 0; while(f1 != f2) { if(deep[f1] < deep[f2]) { swap(f1,f2); swap(u,v); } tmp = max(tmp,query(1,p[f1],p[u])); u = fa[f1]; f1 = top[u]; } if(u == v)return tmp; if(deep[u] > deep[v]) swap(u,v); return max(tmp,query(1,p[son[u]],p[v])); } int e[MAXN][3]; int main() { int T; int n; scanf("%d",&T); while(T--) { init(); scanf("%d",&n); for(int i = 0;i < n-1;i++) { scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]); addedge(e[i][0],e[i][1]); addedge(e[i][1],e[i][0]); } dfs1(1,0,0); getpos(1,1); build(1,0,pos-1); for(int i = 0;i < n-1; 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 op[10]; int u,v; while(scanf("%s",op) == 1) { if(op[0] == 'D')break; scanf("%d%d",&u,&v); if(op[0] == 'Q') printf("%d ",find(u,v)); else update(1,p[e[u-1][1]],v); } } return 0; }