poj 1744 tree 树分治
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Tree
Time Limit: 1000MS | Memory Limit: 30000K | |
Description
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.
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
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.
The last test case is followed by two zeros.
Output
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 0
Sample Output
8
Source
树分治模板题;
分治算法在树的路径问题中的应用
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define LL long long #define pi (4*atan(1.0)) #define eps 1e-8 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e4+10,M=2e6+10,inf=1e9+10; const LL INF=1e18+10,mod=1e9+7; struct is { int v,w,nex; }edge[N<<1]; int head[N],edg; void add(int u,int v,int w) { edge[++edg]=(is){v,w,head[u]};head[u]=edg; } int son[N],msi[N],d[N]; int vis[N],deep[N]; int n,K,ans,root,sum; void groot(int u,int fa) { son[u]=1,msi[u]=0; for(int i=head[u];i!=-1;i=edge[i].nex) { int v=edge[i].v; if(v==fa||vis[v])continue; groot(v,u); son[u]+=son[v]; msi[u]=max(msi[u],son[v]); } msi[u]=max(msi[u],sum-son[u]); if(msi[u]<msi[root])root=u; } void gdeep(int x,int fa) { deep[++deep[0]]=d[x]; for(int i=head[x];i!=-1;i=edge[i].nex) { int v=edge[i].v; int w=edge[i].w; if(v==fa||vis[v])continue; d[v]=d[x]+w; gdeep(v,x); } } int rootans(int x,int base) { d[x]=base;deep[0]=0; gdeep(x,0); sort(deep+1,deep+1+deep[0]); int ans=0,l=1,r=deep[0]; while(l<r) { if(deep[l]+deep[r]<=K) { ans+=r-l; l++; } else r--; } return ans; } void dfs(int u) { ans+=rootans(u,0); vis[u]=1; for(int i=head[u];i!=-1;i=edge[i].nex) { int v=edge[i].v; int w=edge[i].w; if(vis[v])continue; ans-=rootans(v,w); root=0;sum=son[v]; groot(v,u); dfs(root); } } void init(int n) { memset(head,-1,sizeof(head)); memset(msi,0,sizeof(msi)); memset(son,0,sizeof(son)); memset(d,0,sizeof(d)); memset(vis,0,sizeof(vis)); memset(deep,0,sizeof(deep)); sum=n;root=0;edg=0; msi[0]=inf; ans=0; } int main() { while(~scanf("%d%d",&n,&K)) { if(n==0&&K==0)break; init(n); for(int i=1;i<n;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } groot(1,0); dfs(root); printf("%d\n",ans); } return 0; }
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