1468: Tree
Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 1023 Solved: 532
[Submit][Status][Discuss]
Description
给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K
Input
N（n<=40000） 接下来n-1行边描述管道，按照题目中写的输入 接下来是k
Output
一行，有多少对点之间的距离小于等于k
Sample Input
7
1 6 13
6 3 9
3 5 7
4 1 3
2 4 20
4 7 2
10
1 6 13
6 3 9
3 5 7
4 1 3
2 4 20
4 7 2
10
Sample Output
5
HINT
Source
题解：
点分治裸题，那么什么是点分治呢
点分治
1、求树的重心
2、计算以当前重心为根的子树的答案
3、去掉以当前重心儿子为根的子树的答案
4、枚举每个儿子，分治
代码：
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cmath> 6 #define maxn 40005 7 using namespace std; 8 int n,k,i,j,tot,root,ans,L,all; 9 int pre[maxn*2],now[maxn],v[maxn*2],val[maxn*2],son[maxn],f[maxn]; 10 int dist[maxn],data[maxn]; 11 bool vis[maxn]; 12 int read() 13 { 14 int x=0; char ch; bool bo=0; 15 while (ch=getchar(),ch<'0'||ch>'9') if (ch=='-') bo=1; 16 while (x=x*10+ch-'0',ch=getchar(),ch>='0'&&ch<='9'); 17 if (bo) return -x; return x; 18 } 19 void ins(int a,int b,int c){ 20 pre[++tot]=now[a]; now[a]=tot; v[tot]=b; val[tot]=c; 21 } 22 void get_root(int u,int fa) 23 { 24 son[u]=1; f[u]=0; 25 for (int p=now[u]; p; p=pre[p]) 26 { 27 int vv=v[p]; 28 if (vv==fa || vis[vv]) continue; 29 get_root(vv,u); son[u]+=son[vv]; 30 f[u]=max(f[u],son[vv]); 31 } 32 f[u]=max(f[u],all-son[u]); 33 if (f[u]<f[root]) root=u; 34 } 35 void get_array(int u,int fa) 36 { 37 data[++L]=dist[u]; 38 for (int p=now[u]; p; p=pre[p]) 39 { 40 int vv=v[p]; 41 if (vv!=fa&&!vis[vv]) 42 dist[vv]=dist[u]+val[p],get_array(vv,u); 43 } 44 } 45 int calc(int u,int value) 46 { 47 dist[u]=value; L=0; get_array(u,-1); 48 sort(data+1,data+1+L); 49 int preans=0,l=1,r=L; 50 while (l<r) 51 { 52 if (data[l]+data[r]<=k) preans+=(r-l),l++; else r--; 53 } 54 return preans; 55 } 56 void solve(int u) 57 { 58 ans+=calc(u,0); vis[u]=1; 59 //cout<<" ans "<<u<<endl; 60 for (int p=now[u]; p; p=pre[p]) 61 { 62 int vv=v[p]; 63 //cout<<" "<<vv<<endl; 64 if (vis[vv]) continue; 65 ans-=calc(vv,val[p]); f[root=0]=n+1; 66 all=son[vv]; get_root(vv,-1); 67 solve(root); 68 } 69 } 70 int main() 71 { 72 n=read(); 73 for (int i=1; i<=n-1; i++) 74 { 75 int u=read(),v=read(),value=read(); 76 ins(u,v,value); ins(v,u,value); 77 } 78 k=read(); 79 f[root=0]=n+1; all=n; 80 get_root(1,-1); 81 //cout<<" "<<root<<endl; 82 solve(root); 83 printf("%d ",ans); 84 return 0; 85 }