首页 > 代码库 > SPOJ 375. Query on a tree (树链剖分)
SPOJ 375. Query on a tree (树链剖分)
Query on a tree
5000ms
262144KB
This problem will be judged on SPOJ. Original ID: QTREE
64-bit integer IO format: %lld Java class name: Main
64-bit integer IO format: %lld Java class name: Main
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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:131 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONEOutput:13
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstdlib> 4 #include<cstring> 5 using namespace std; 6 const int maxn = 1e4+13; 7 8 struct node 9 { 10 int l,r,Max; 11 }f[maxn*4]; 12 13 struct Edge 14 { 15 int to,next; 16 }edge[maxn*2]; 17 18 int e[maxn][3]; 19 int p[maxn]; 20 int top[maxn]; 21 int siz[maxn]; 22 int son[maxn]; 23 int deep[maxn]; 24 int father[maxn]; 25 int head[maxn]; 26 int num[maxn]; 27 int cont,pos; 28 29 void init() 30 { 31 cont = 0; 32 pos = 0; 33 memset(head,-1,sizeof(head)); 34 memset(son,-1,sizeof(son)); 35 } 36 void add(int n1,int n2) 37 { 38 edge[cont].to=n2;// 指向谁 39 edge[cont].next=head[n1]; 40 head[n1]=cont; 41 cont++; 42 } 43 44 void dfs1(int u,int pre,int d)/**fa deep,num,son**/ 45 { 46 deep[u]=d; 47 father[u]=pre; 48 num[u]=1; 49 for(int i=head[u];i!=-1;i=edge[i].next) 50 { 51 int v = edge[i].to; 52 if(v!=pre) 53 { 54 dfs1(v,u,d+1); 55 num[u]=num[u]+num[v]; 56 if(son[u]==-1 || num[v]>num[son[u]]) 57 son[u]=v; 58 } 59 } 60 } 61 62 void getops(int u,int sp) 63 { 64 top[u]=sp; 65 if(son[u]!=-1) 66 { 67 p[u]=++pos; 68 getops(son[u],sp); 69 } 70 else 71 { 72 p[u]=++pos; 73 return; 74 } 75 for(int i=head[u];i!=-1;i=edge[i].next) 76 { 77 int v = edge[i].to; 78 if(v!=son[u] && v!=father[u]) 79 getops(v,v); 80 } 81 } 82 void build(int l,int r,int n) 83 { 84 f[n].l=l; 85 f[n].r=r; 86 f[n].Max=0; 87 if(l==r)return; 88 int mid=(l+r)/2; 89 build(l,mid,n*2); 90 build(mid+1,r,n*2+1); 91 } 92 int query(int l,int r,int n) 93 { 94 int mid=(f[n].l+f[n].r)/2; 95 int ans1,ans2; 96 if(f[n].l==l && f[n].r==r) return f[n].Max; 97 if(mid>=r) return query(l,r,n*2); 98 else if(mid<l) return query(l,r,n*2+1); 99 else100 {101 ans1=query(l,mid,n*2);102 ans2=query(mid+1,r,n*2+1);103 if(ans1<ans2) ans1=ans2;104 }105 return ans1;106 }107 void update(int x,int num1,int n)108 {109 int mid=(f[n].l+f[n].r)/2;110 if(f[n].l == x && f[n].r == x)111 {112 f[n].Max=num1;113 return;114 }115 if(mid>=x) update(x,num1,n*2);116 else update(x,num1,n*2+1);117 f[n].Max = f[n*2].Max>f[n*2+1].Max? f[n*2].Max:f[n*2+1].Max;118 }119 int find(int u,int v)120 {121 int f1 = top[u],f2 = top[v];122 int MAX=0;123 while(f1!=f2)124 {125 if(deep[f1]<deep[f2])126 {127 swap(f1,f2);128 swap(u,v);129 }130 MAX=max(MAX,query(p[f1],p[u],1));131 u=father[f1];132 f1=top[u];133 }134 if(u==v)return MAX;135 if(deep[u]>deep[v])swap(u,v);136 return max(MAX,query(p[son[u]],p[v],1));137 }138 int main()139 {140 int T,n,l,r,x,num1;141 char a[12];142 scanf("%d",&T);143 while(T--)144 {145 scanf("%d",&n);146 init();147 for(int i=1;i<n;i++)148 {149 scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);150 add(e[i][0],e[i][1]);151 add(e[i][1],e[i][0]);152 }153 dfs1(1,0,0);154 getops(1,1);155 build(1,pos,1);156 for(int i=1;i<n;i++)157 {158 if(deep[e[i][0]]>deep[e[i][1]]) swap(e[i][0],e[i][1]);159 update(p[e[i][1]],e[i][2],1);160 }161 while(scanf("%s",a)>0)162 {163 if(a[0]==‘D‘)break;164 if(a[0]==‘Q‘)165 {166 scanf("%d%d",&l,&r);167 printf("%d\n",find(l,r));168 }169 else if(a[0]==‘C‘)170 {171 scanf("%d%d",&x,&num1);172 update(p[e[x][1]],num1,1);173 }174 }175 }176 return 0;177 }
SPOJ 375. Query on a tree (树链剖分)
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