题目描述
在一个点数为N(N<10,000)的带权树上,支持两个操作:1. 改变一个边权 2. 询问u和v之间的路径上的最大边权
吐槽:
1. 这就是树链剖分最基本的应用了吧,写得我要死了....
2. 常数好大... 2.9s
算法分析:
就是树的轻重边剖分了。轻边直接修改,重边因为是一段一段连续的,所以用一颗线段树维护起来....
具体怎么实现的明天再说....
查询两个点u和v就是求max(path(u,lca(u,v)),path(v,lca(u,v)))。
lca用的是clj版的....
查询就是重边用RMQ,轻边直接查询。
明早起来再详细讲....实验室要关门了...
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cassert>
5 using namespace std;
6 template <typename T> inline void chkmax(T &a, T b){ if(a < b) a = b;}
7 // build tree
8 const int V = 10005;
9 const int E = V*2;
10 int head[V], nxt[E], pnt[E], e, n, cost[E];
11 void add_edge(int u,int v,int c){
12 nxt[e] = head[u];
13 head[u] = e;
14 pnt[e] = v;
15 cost[e] = c;
16 e++;
17 }
18 // segment tree & prepare
19 const int inf = ~0u>>2;
20 int seg[E<<2], parent[V], M, UP[V], heavy[V], num[E], deep[V], P[V], sz[V], segsz;
21 void ins(int pos,int val){
22 pos += M;
23 seg[pos] = val;
24 while(pos>>=1) {
25 seg[pos] = max(seg[pos<<1], seg[pos<<1|1]);
26 }
27 }
28 int find(int u){ return u==parent[u] ? u: parent[u] = find(parent[u]);}
29 int rmq(int l,int r){
30 int ans = -inf;
31 for(l += M-1, r += M+1; l^r^1; l>>=1, r>>=1){
32 if(~l&1) chkmax(ans,seg[l^1]);
33 if(r&1) chkmax(ans,seg[r^1]);
34 }
35 return ans;
36 }
37 void dfs(int u,int f){
38 int mx = 0 ,s = -1;
39 sz [u] = 1;
40 for(int i= head[u]; i!=-1; i=nxt[i]){
41 int v = pnt[i];
42 if(v == f) continue;
43 P[v] = i^1;
44 deep[v] = deep[u]+1;
45 dfs(v,u);
46 sz[u] += sz[v];
47 if(sz[v] > mx){
48 mx = sz[v];
49 s = i;
50 }
51 }
52 heavy[u] = s;
53 if(s != -1){
54 parent[pnt[s]] = u;
55 }
56 }
57 void prepare(){
58 for(int i =30; i;i--) if((1<<i) > n) M = 1<<i;
59 for(int i = M; i<2*M; i++) seg[i] = -inf;
60 memset(num,-1,sizeof(num));
61 segsz = deep[0] = 0;
62 for(int i=0;i<n;i++) parent[i] = i;
63 dfs(0,0);
64 // for(int i=0;i<n;i++) cout<<deep[i]<<" "; cout<<endl;
65 for(int i=0; i<n; i++) if(heavy[i] == -1){
66 int pos = i;
67 // cout<<"i:"<<i<<endl;
68 while(pos && pnt[heavy[pnt[P[pos]]]] == pos){
69 int t = P[pos];
70 // cout<<"pos : "<<pos<<" "<<t<<endl;
71 num[t^1] = num[t] = ++segsz;
72 ins(segsz,cost[t]);
73 pos = pnt[P[pos]];
74 //cout<<pos<<" "<<pnt[P[pos]]<<" "<<pnt[heavy[pnt[P[pos]]]]<<endl;
75 }
76 }
77 // for(int i=0;i<e;i++) cout<<num[i]<<" ";cout<<endl;
78 }
79 // operator
80 void change(int pos,int val){
81 if(num[pos] == -1) { cost[pos] = cost[pos^1] = val; return;}
82 ins(num[pos],val);
83 }
84 int query(int u,int v){
85 //cout<<u<<" "<<v<<endl;
86 int ans = -inf;
87 while(u!=v){
88 int r = P[u];
89 if(num[r] == -1){
90 chkmax(ans,cost[r]);
91 u = pnt[r];
92 }
93 else {
94 int p = parent[u];
95 if(deep[p] < deep[v]) p = v;
96 u = p;
97 //cout<<r<<" "<<heavy[p]<<endl;
98 int l = num[heavy[p]]; r = num[r];
99 // cout<<"query : "<<l<<" "<<r<<endl;
100 assert(l>0 && l>=r);
101 int t = rmq(r,l);
102 chkmax(ans,t);
103 }
104 }
105 return ans;
106 }
107 int lca(int u,int v){
108 while(1){
109 int a = find(u) , b = find(v);
110 if(a == b) return deep[u]< deep[v] ? u : v;
111 else if(deep[a] >= deep[b]) u = pnt[P[a]]; else v = pnt[P[b]];
112 }
113 }
114 int ask(int u,int v){
115 int p = lca(u,v);
116 return max(query( u, p),query( v, p));
117 }
118 int main(){
119 int t; cin >> t;
120 while(t--){
121 int u,v,c; scanf("%d",&n);
122 e = 0; memset(head,-1,sizeof(head));
123 for(int i=1;i<n;i++){
124 scanf("%d%d%d",&u,&v,&c);
125 u--;v--;
126 add_edge(u,v,c);
127 add_edge(v,u,c);
128 }
129 prepare();
130 char ch[20];
131 while(scanf("%s",ch) && strcmp(ch,"DONE")){
132 if(ch[0]=='C'){
133 scanf("%d%d",&u,&c);
134 change((u-1)*2,c);
135 }
136 else {
137 scanf("%d%d",&u,&v);
138 printf("%d\n",ask(u-1,v-1));
139 }
140 }
141 }
142 }
143
posted on 2012-05-14 22:17
西月弦 阅读(809)
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