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poj 1741(树的点分治)
Tree
Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
The input contains several test cases. The first line of each
test case contains two integers n, k. (n<=10000) The following n-1
lines each contains three integers u,v,l, which means there is an edge
between node u and v of length l.
The last test case is followed by two zeros.
Output
The last test case is followed by two zeros.
For each test case output the answer on a single line.
Sample Input
5 4 1 2 3 1 3 1 1 4 2 3 5 1 0 0Sample Output
8
树的点分治 参考了漆子超的论文 以及多位大佬的博客
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<queue> 7 #include<map> 8 #include<set> 9 #include<vector> 10 #include<cstdlib> 11 #include<stack> 12 #include<string> 13 #define eps 0.000000001 14 typedef long long ll; 15 typedef unsigned long long LL; 16 using namespace std; 17 const int INF=0x3f3f3f3f; 18 const int N=100000+10; 19 int son[N]; 20 int ans; 21 int num; 22 int p; 23 int head[N]; 24 int vis[N]; 25 int minn; 26 int root; 27 int k; 28 int dis[N]; 29 void init(){ 30 num=0; 31 memset(vis,0,sizeof(vis)); 32 memset(head,-1,sizeof(head)); 33 } 34 struct node{ 35 int to,next,w; 36 }edge[N]; 37 void add(int u,int v,int w){ 38 edge[num].to=v; 39 edge[num].w=w; 40 edge[num].next=head[u]; 41 head[u]=num++; 42 } 43 void getroot(int x,int y,int n){ 44 son[x]=1; 45 int maxx=0; 46 for(int i=head[x];i!=-1;i=edge[i].next){ 47 int v=edge[i].to; 48 if(v==y||vis[v])continue; 49 getroot(v,x,n); 50 son[x]=son[x]+son[v]; 51 maxx=max(maxx,son[v]); 52 } 53 maxx=max(maxx,n-son[x]); 54 //cout<<maxx<<" "<<x<<endl; 55 if(minn>maxx){ 56 minn=maxx; 57 root=x; 58 } 59 } 60 void getdeep(int x,int v0,int y){ 61 dis[p++]=y; 62 for(int i=head[x];i!=-1;i=edge[i].next){ 63 int v=edge[i].to; 64 if(v==v0||vis[v])continue; 65 getdeep(v,x,y+edge[i].w); 66 } 67 } 68 int cal(int x,int y){ 69 int res=0; 70 p=0; 71 getdeep(x,0,y); 72 sort(dis,dis+p); 73 int l=0; 74 int r=p-1; 75 while(l<r){ 76 if(dis[l]+dis[r]<=k){ 77 res=res+(r-l); 78 l++; 79 } 80 else{ 81 r--; 82 } 83 } 84 return res; 85 } 86 void work(int x){ 87 ans=ans+cal(x,0); 88 //cout<<cal(x,0)<<" "; 89 // cout<<endl; 90 vis[x]=1; 91 for(int i=head[x];i!=-1;i=edge[i].next){ 92 int v=edge[i].to; 93 if(vis[v]==0){ 94 ans=ans-cal(v,edge[i].w); 95 minn=INF; 96 getroot(v,0,son[v]); 97 work(root); 98 } 99 } 100 } 101 int main(){ 102 int n; 103 while(scanf("%d%d",&n,&k)!=EOF){ 104 if(n==0&&k==0)break; 105 init(); 106 for(int i=0;i<n-1;i++){ 107 int u,v,w; 108 scanf("%d%d%d",&u,&v,&w); 109 add(u,v,w); 110 add(v,u,w); 111 } 112 ans=0; 113 minn=INF; 114 getroot(1,-1,n); 115 //for(int i=1;i<=n;i++)cout<<son[i]<<" "; 116 // cout<<endl; 117 //cout<<root<<" "<<minn<<endl; 118 work(root); 119 cout<<ans<<endl; 120 } 121 return 0; 122 }
poj 1741(树的点分治)
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