Query on a tree
Time Limit: 5000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: QTREE64-bit integer IO format: %lld Java class name: Main
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
解题：树链剖分
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 10100; 4 struct arc{ 5 int to,next; 6 arc(int x = 0,int y = -1){ 7 to = x; 8 next = y; 9 } 10 }e[30000]; 11 struct node{ 12 int lt,rt,val; 13 }tree[maxn<<2]; 14 int head[maxn],fa[maxn],dep[maxn],son[maxn],tot; 15 int siz[maxn],top[maxn],tid[maxn],label; 16 void add(int u,int v){ 17 e[tot] = arc(v,head[u]); 18 head[u] = tot++; 19 } 20 void Find_heavy_edge(int u,int father,int depth){ 21 siz[u] = 1; 22 dep[u] = depth; 23 fa[u] = father; 24 son[u] = -1; 25 for(int i = head[u]; ~i; i = e[i].next){ 26 if(e[i].to == father) continue; 27 Find_heavy_edge(e[i].to,u,depth + 1); 28 siz[u] += siz[e[i].to]; 29 if(son[u] == -1 || siz[e[i].to] > siz[son[u]]) 30 son[u] = e[i].to; 31 } 32 } 33 void connect_heavy_edge(int u,int ancestor){ 34 tid[u] = label++; 35 top[u] = ancestor; 36 if(son[u] != -1) connect_heavy_edge(son[u],ancestor); 37 for(int i = head[u]; ~i; i = e[i].next){ 38 if(e[i].to == fa[u] || e[i].to == son[u]) continue; 39 connect_heavy_edge(e[i].to,e[i].to); 40 } 41 } 42 inline void pushup(int v){ 43 tree[v].val = max(tree[v<<1].val,tree[v<<1|1].val); 44 } 45 void build(int lt,int rt,int v){ 46 tree[v].lt = lt; 47 tree[v].rt = rt; 48 tree[v].val = 0; 49 if(lt == rt) return; 50 int mid = (lt + rt)>>1; 51 build(lt,mid,v<<1); 52 build(mid+1,rt,v<<1|1); 53 } 54 void update(int pos,int val,int v){ 55 if(tree[v].lt == tree[v].rt){ 56 tree[v].val = val; 57 return; 58 } 59 if(pos <= tree[v<<1].rt) update(pos,val,v<<1); 60 if(pos >= tree[v<<1|1].lt) update(pos,val,v<<1|1); 61 pushup(v); 62 } 63 int query(int lt,int rt,int v){ 64 int ret = 0; 65 if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].val; 66 if(lt <= tree[v<<1].rt) ret = query(lt,rt,v<<1); 67 if(rt >= tree[v<<1|1].lt) ret = max(ret,query(lt,rt,v<<1|1)); 68 return ret; 69 } 70 int Find(int u,int v){ 71 int f1 = top[u],f2 = top[v],tmp = 0; 72 while(f1 != f2){ 73 if(dep[f1] < dep[f2]){ 74 swap(f1,f2); 75 swap(u,v); 76 } 77 tmp = max(tmp,query(tid[f1],tid[u],1)); 78 u = fa[f1]; 79 f1 = top[u]; 80 } 81 if(u == v) return tmp; 82 if(dep[u] > dep[v]) swap(u,v); 83 return max(tmp,query(tid[son[u]],tid[v],1)); 84 } 85 int ac[maxn][3]; 86 int main(){ 87 int kase,n,u,v; 88 char op[10]; 89 scanf("%d",&kase); 90 while(kase--){ 91 memset(head,-1,sizeof head); 92 scanf("%d",&n); 93 for(int i = tot = 0; i < n-1; ++i){ 94 scanf("%d%d%d",&ac[i][0],&ac[i][1],&ac[i][2]); 95 add(ac[i][0],ac[i][1]); 96 add(ac[i][1],ac[i][0]); 97 } 98 label = 0; 99 Find_heavy_edge(1,0,0); 100 connect_heavy_edge(1,1); 101 build(0,label-1,1); 102 for(int i = 0; i < n-1; ++i){ 103 if(dep[ac[i][0]] > dep[ac[i][1]]) 104 swap(ac[i][0],ac[i][1]); 105 update(tid[ac[i][1]],ac[i][2],1); 106 } 107 while(~scanf("%s",op)){ 108 if(op[0] == 'D') break; 109 scanf("%d%d",&u,&v); 110 if(op[0] == 'Q') printf("%d ",Find(u,v)); 111 else update(tid[ac[u-1][1]],v,1); 112 } 113 } 114 return 0; 115 }
Link-Cut Tree
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 10010; 4 struct arc{ 5 int to,w,next; 6 arc(int x = 0,int y = 0,int z = -1){ 7 to = x; 8 w = y; 9 next = z; 10 } 11 }e[maxn<<1]; 12 int head[maxn],tot; 13 void add(int u,int v,int w){ 14 e[tot] = arc(v,w,head[u]); 15 head[u] = tot++; 16 } 17 struct LCT{ 18 int fa[maxn],pa[maxn],ch[maxn][2],sz[maxn],maxv[maxn],belong[maxn],key[maxn]; 19 inline void pushup(int x){ 20 maxv[x] = max(max(maxv[ch[x][0]],maxv[ch[x][1]]),key[x]); 21 } 22 void rotate(int x,int kd){ 23 int y = fa[x]; 24 ch[y][kd^1] = ch[x][kd]; 25 fa[ch[x][kd]] = y; 26 fa[x] = fa[y]; 27 fa[y] = x; 28 ch[x][kd] = y; 29 if(fa[x]) ch[fa[x]][y == ch[fa[x]][1]] = x; 30 pushup(y); 31 } 32 void splay(int x,int goal = 0){ 33 while(fa[x] != goal){ 34 if(fa[fa[x]] == goal) rotate(x,x == ch[fa[x]][0]); 35 else{ 36 int y = fa[x],z = fa[y],s = (y == ch[z][0]); 37 if(x == ch[y][s]){ 38 rotate(x,s^1); 39 rotate(x,s); 40 }else{ 41 rotate(y,s); 42 rotate(x,s); 43 } 44 } 45 } 46 pushup(x); 47 } 48 void access(int x){ 49 for(int y = 0; x; x = pa[x]){ 50 splay(x); 51 fa[ch[x][1]] = 0; 52 pa[ch[x][1]] = x; 53 ch[x][1] = y; 54 fa[y] = x; 55 pa[y] = 0; 56 y = x; 57 pushup(x); 58 } 59 } 60 void build(int _val,int parent,int x){ 61 pa[x] = parent; 62 maxv[x] = key[x] = _val; 63 ch[x][0] = ch[x][1] = fa[x] = 0; 64 } 65 void change(int ith,int val){ 66 int x = belong[ith-1]; 67 key[x] = val; 68 splay(x); 69 } 70 int query(int x,int y){ 71 access(y); 72 for(y = 0; x; x = pa[x]){ 73 splay(x); 74 if(!pa[x]) return max(maxv[y],maxv[ch[x][1]]); 75 fa[ch[x][1]] = 0; 76 pa[ch[x][1]] = x; 77 ch[x][1] = y; 78 fa[y] = x; 79 pa[y] = 0; 80 y = x; 81 pushup(x); 82 } 83 return 0; 84 } 85 }lct; 86 void dfs(int u,int fa){ 87 for(int i = head[u]; ~i; i = e[i].next){ 88 if(e[i].to == fa) continue; 89 lct.build(e[i].w,u,e[i].to); 90 lct.belong[i>>1] = e[i].to; 91 dfs(e[i].to,u); 92 } 93 } 94 int main(){ 95 int kase,n,u,v,w; 96 char op[10]; 97 scanf("%d",&kase); 98 while(kase--){ 99 memset(head,-1,sizeof head); 100 tot = 0; 101 scanf("%d",&n); 102 for(int i = 1; i < n; ++i){ 103 scanf("%d%d%d",&u,&v,&w); 104 add(u,v,w); 105 add(v,u,w); 106 } 107 dfs(1,1); 108 while(~scanf("%s",op)){ 109 if(op[0] == 'D') break; 110 scanf("%d%d",&u,&v); 111 if(op[0] == 'Q') printf("%d ",lct.query(u,v)); 112 else if(op[0] == 'C') lct.change(u,v); 113 } 114 } 115 return 0; 116 }