动态树(Link Cut Tree) :SPOJ 375 Query on a tree
QTREE - Query on a tree
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
这题可以用树链剖分做,我这里用LCT做的,代码量更少。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 const int maxn=10010; 7 int Max[maxn],fa[maxn],ch[maxn][2],key[maxn]; 8 bool rt[maxn]; 9 10 void Push_up(int p) 11 { 12 Max[p]=max(key[p],max(Max[ch[p][0]],Max[ch[p][1]])); 13 } 14 15 void Rotate(int x) 16 { 17 int y=fa[x],g=fa[y],c=ch[y][1]==x; 18 ch[y][c]=ch[x][c^1];ch[x][c^1]=y; 19 fa[ch[y][c]]=y;fa[y]=x;fa[x]=g; 20 if(rt[y]) 21 rt[y]=false,rt[x]=true; 22 else 23 ch[g][ch[g][1]==y]=x; 24 Push_up(y); 25 } 26 27 void Splay(int x) 28 { 29 for(int y=fa[x];!rt[x];Rotate(x),y=fa[x]) 30 if(!rt[y]) 31 Rotate((ch[fa[y]][1]==y)==(ch[y][1]==x)?y:x); 32 Push_up(x); 33 } 34 35 void Access(int x) 36 { 37 int y=0; 38 while(x){ 39 Splay(x); 40 rt[ch[x][1]]=true; 41 rt[ch[x][1]=y]=false; 42 Push_up(x); 43 x=fa[y=x]; 44 } 45 } 46 47 void Query(int x,int y) 48 { 49 Access(y),y=0; 50 while(true) 51 { 52 Splay(x); 53 if(!fa[x]){ 54 printf("%d\n",max(Max[y],Max[ch[x][1]])); 55 return; 56 } 57 rt[ch[x][1]]=true; 58 rt[ch[x][1]=y]=false; 59 Push_up(x); 60 x=fa[y=x]; 61 } 62 } 63 64 void Change(int x,int d) 65 { 66 Access(x); 67 Splay(x); 68 key[x]=d; 69 Push_up(x); 70 } 71 72 int fir[maxn],nxt[maxn<<1],to[maxn<<1],cnt; 73 int e[maxn][3]; 74 75 void addedge(int a,int b) 76 { 77 nxt[++cnt]=fir[a];to[cnt]=b;fir[a]=cnt; 78 } 79 80 void DFS(int node) 81 { 82 for(int i=fir[node];i;i=nxt[i]) 83 { 84 if(fa[to[i]])continue; 85 fa[to[i]]=node; 86 DFS(to[i]); 87 } 88 } 89 90 void Init() 91 { 92 cnt=0;Max[0]=-1000000000; 93 for(int i=1;i<=10000;i++){ 94 Max[i]=fir[i]=fa[i]=0; 95 rt[i]=1; 96 } 97 return; 98 } 99 int main() 100 { 101 int T,n,a,b; 102 scanf("%d",&T); 103 while(T--) 104 { 105 Init(); 106 scanf("%d",&n); 107 for(int i=1;i<n;i++) 108 scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]); 109 for(int i=1;i<n;i++) 110 addedge(e[i][0],e[i][1]),addedge(e[i][1],e[i][0]); 111 fa[1]=-1; 112 DFS(1); 113 fa[1]=0; 114 for(int i=1;i<n;i++) 115 { 116 if(fa[e[i][0]]==e[i][1]) 117 swap(e[i][0],e[i][1]); 118 Change(e[i][1],e[i][2]); 119 } 120 char op[10]; 121 while(true) 122 { 123 scanf("%s",op); 124 if(!strcmp(op,"DONE"))break; 125 else if(!strcmp(op,"QUERY")){ 126 scanf("%d%d",&a,&b); 127 Query(a,b); 128 } 129 130 else { 131 scanf("%d%d",&a,&b); 132 Change(e[a][1],b); 133 } 134 } 135 } 136 137 return 0; 138 }
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