A - Tree POJ - 1741
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
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
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
点分治模板题
#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 10007;
struct edge{
int v,w;
};
bool vis[N];
vector<edge> G[N];
int n,k;
int root,son[N],ms[N],sz;
long long ret = 0;
vector<int> vp;
void getroot(int x,int f){
ms[x] = 0;
son[x] = 1;
for(int i = 0;i < G[x].size();i ++){
edge &e = G[x][i];
if(e.v == f || vis[e.v]==false) continue;
getroot(e.v,x);
son[x] += son[e.v];
ms[x] = max(ms[x],son[e.v]);
}
ms[x] = max(ms[x],sz-son[x]);
if(ms[x] < ms[root]) root = x;
//cout <<"!!!" << x << ' '<< ms[x] << ' '<<sz << endl;
}
void getdep(int x,int d,int f){
//cout << x << ' '<<d << ' ' << f << endl;
son[x] = 1;
if(d <= k)vp.push_back(d);
for(int i = 0;i < G[x].size();i ++){
int v = G[x][i].v,w = G[x][i].w;
if(v == f||vis[v]==false) continue;
getdep(v,d+w,x);
son[x] += son[v];
}
}
void dfs(int x,int f){
vp.clear();
long long ans = 0;
for(int i = 0;i < G[root].size();i ++){
int v = G[root][i].v;
int tmp = vp.size();
if(v == f||vis[v] == false) continue;
getdep(v,G[root][i].w,root);
sort(vp.begin()+tmp,vp.end());
int j = vp.size()-1;
for(int t = tmp;t < vp.size();t ++){
while(j >= tmp && vp[j]+vp[t] > k) j --;
if(j >= t) ans -= j-tmp;
else ans -= j-tmp+1;
}
}
sort(vp.begin(),vp.end());
int j = vp.size()-1;
for(int i = 0;i < vp.size();i ++){
while(j >= 0 && vp[j]+vp[i] > k) j --;
if(j >= i) ans += j;
else ans += j+1;
//cout << "###" << i << ' '<<j << ' '<< vp[i]<< ' '<<vp[j] << endl;
}
//cout << "!!!!"<< x << ' '<<root << ' ' << tmp << ' '<< vp.size()<<endl;
ans /= 2;
ans += vp.size();
ret += ans;
int rt = root;
vis[root] = false;
//cout << root << ' ' << ans << endl;
for(int i = 0;i < G[rt].size();i ++){
int v = G[rt][i].v;
if(v == f||vis[v]==false) continue;
sz = son[v];
root = 0;
getroot(v,rt);
dfs(v,rt);
}
}
int main(){
while(scanf("%d %d",&n,&k)==2){
if(n == 0) break;
memset(vis,true,sizeof vis);
ret = 0;
for(int i = 1;i <= n;i ++) G[i].clear();
for(int i = 1;i < n;i ++) {
int u,v,w;
scanf("%d %d %d",&u,&v,&w);
G[u].push_back({v,w});
G[v].push_back({u,w});
}
root = 0;
ms[0] = n;
sz = n;
getroot(1,-1);
dfs(1,-1);
cout << ret << endl;
}
return 0;
}