| Time Limit: 851MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
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.
Example
Input: 1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE Output: 1 3
【思路】
树链剖分。
划分轻重链，线段树维护。
这里有个知识入门：http://blog.sina.com.cn/s/blog_6974c8b20100zc61.html
【代码】
1 #include<cstdio> 2 #include<vector> 3 #include<cstring> 4 #include<algorithm> 5 #include<iostream> 6 using namespace std; 7 8 const int N = 20000+10; 9 10 struct Edge { int u,v,w; }; 11 vector<int> G[N]; 12 vector<Edge> es; 13 int n,z,root,d[N][3]; 14 int fa[N],siz[N],dep[N],son[N],w[N],top[N]; 15 //fa为父节点 siz为子树大小 dep为节点深度 16 //son代表重儿子 w为u与fa[u]在线段树中的位置 top代表所属重链的顶端 17 //z为线段树大小 18 19 void adde(int u,int v,int w) { 20 es.push_back((Edge){u,v,w}); 21 int m=es.size(); 22 G[u].push_back(m-1); 23 } 24 25 void dfs(int u) { //->siz[] son[] fa[] 26 siz[u]=1; son[u]=0; 27 for(int i=0;i<G[u].size();i++) { 28 int v=es[G[u][i]].v; 29 if(v!=fa[u]) { 30 fa[v]=u; 31 dep[v]=dep[u]+1; 32 dfs(v); 33 if(siz[v]>siz[son[u]]) son[u]=v; 34 siz[u]+=siz[v]; 35 } 36 } 37 } 38 void build_tree(int u,int tp) { //son[] -> top[] w[] 39 w[u]=++z; top[u]=tp; 40 if(son[u]) build_tree(son[u],top[u]); 41 for(int i=0;i<G[u].size();i++) { 42 int v=es[G[u][i]].v; 43 if(v!=son[u] && v!=fa[u]) build_tree(v,v); 44 } 45 } 46 47 int tree[N]; 48 void update(int u,int L,int R,int loc,int x) { 49 if(loc<L || R<loc) return ; 50 if(L==R) { tree[u]=x; return ; } 51 int M=L+(R-L)/2 , lc=u*2,rc=lc+1; 52 update(lc,L,M,loc,x); 53 update(rc,M+1,R,loc,x); 54 tree[u]=max(tree[lc],tree[rc]); 55 } 56 int query(int u,int L,int R,int l,int r) { 57 if(R<l || L>r) return 0; 58 if(l<=L && R<=r) return tree[u]; 59 int M=L+(R-L)/2; 60 return max(query(u*2,L,M,l,r),query(u*2+1,M+1,R,l,r)); 61 } 62 int find(int u,int v) { 63 int f1=top[u] , f2=top[v] , ans=0; 64 while(f1!=f2) { //直到移动到同一重链 65 if(dep[f1]<dep[f2]) 66 swap(f1,f2) , swap(u,v); 67 ans=max(ans,query(1,1,z,w[f1],w[u])); //在重链上移动同时统计 68 u=fa[f1] , f1=top[u]; 69 } 70 if(u==v) return ans; 71 if(dep[u]>dep[v]) swap(u,v); 72 return max(ans,query(1,1,z,w[son[u]],w[v])); //uv之间统计 73 } 74 75 void init() { 76 scanf("%d",&n); 77 es.clear(); 78 for(int i=0;i<=n;i++) G[i].clear(); 79 root=(n+1)/2; 80 fa[root]=dep[root]=z=0; 81 memset(siz,0,sizeof(siz)); 82 memset(tree,0,sizeof(tree)); 83 int u,v,c; 84 for(int i=1;i<n;i++) { 85 scanf("%d%d%d",&u,&v,&c); 86 d[i][0]=u , d[i][1]=v , d[i][2]=c; 87 adde(u,v,c) , adde(v,u,c); 88 } 89 dfs(root); 90 build_tree(root,root); 91 for(int i=1;i<n;i++) { 92 if(dep[d[i][0]]>dep[d[i][1]]) swap(d[i][0],d[i][1]); 93 update(1,1,z,w[d[i][1]],d[i][2]); 94 } 95 } 96 void solve() { 97 char s[20]; 98 int u,v; 99 while(scanf("%s",s)==1 && s[0]!='D') { 100 scanf("%d%d",&u,&v); 101 if(s[0]=='Q') printf("%d ",find(u,v)); 102 else update(1,1,z,w[d[u][1]],v); 103 } 104 } 105 106 int main() { 107 int T; 108 scanf("%d",&T); 109 while(T--) { 110 init(); 111 solve(); 112 } 113 return 0; 114 }