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
题目分析 ： 给定一棵树，以及树上边的关系大小，问你又多少对点的距离是小于等于所给定的 d 的
思路分析 ： 树上点分治的板子题，首先寻找树的重心，以重心为根结点，寻求所有符合题意要求的点对，但是这样计算会算出一些不符合题目的点对，在减去即可，此时当遍历到一个新的结点时，此时的情况又可以当成最初的情况，找重心的时候要注意，对它的子树来说，总的结点数是小于 n 的！！！
代码示例 ：
const int maxn = 1e4+5;
const int inf = 0x3f3f3f3f;
#define ll long long
int n, m;
struct node
{
int to, cost;
node(int _to = 0, int _cost = 0):to(_to), cost(_cost){}
};
vector<node>ve[maxn];
int root;
int size[maxn], mx[maxn]; // size表示每个结点所连的结点数， mx表示对每个根结点所连的最大结点子树有多少的结点
int balance;
bool done[maxn];
int ans = 0;
int numm; // 表示结点总数
void getroot(int x, int fa){
size[x] = 1, mx[x] = 0;
for(int i = 0; i < ve[x].size(); i++){
int to = ve[x][i].to;
if (to == fa || done[to]) continue;
getroot(to, x);
size[x] += size[to];
mx[x] = max(mx[x], size[to]);
}
mx[x] = max(mx[x], numm-size[x]); // 对子树在寻找子树的重心的过程中，子树的总结点数是会变小的
if (mx[x] < balance) {balance = mx[x], root = x;}
}
int cnt = 0;
int dep[maxn];
void dfssize(int x, int fa, int d){
dep[cnt++] = d;
for(int i = 0; i < ve[x].size(); i++){
int to = ve[x][i].to;
int cost = ve[x][i].cost;
if (to == fa || done[to]) continue;
dfssize(to, x, d+cost);
}
}
int cal(int x, int d){
cnt = 0;
dfssize(x, x, d);
sort(dep, dep+cnt);
int l = 0, r = cnt-1;
int sum = 0;
while(l < r){
if (dep[l]+dep[r] <= m){
sum += r-l;
l++;
}
else r--;
}
//printf("sum = %d \n", sum);
//system("pause");
return sum;
}
void dfs(int x){
done[x] = true;
ans += cal(x, 0);
for(int i = 0; i < ve[x].size(); i++){
int to = ve[x][i].to;
int cost = ve[x][i].cost;
if (done[to]) continue;
ans -= cal(to, cost);
balance = inf;
numm = size[to]; // 这里是重点，因为这个地方一直T，还以为写的代码有问题
getroot(to, to);
//printf("root = %d\n", root);
dfs(root);
}
}
int a, b, w;
int main() {
while(scanf("%d%d", &n, &m) && n+m){
for(int i = 0; i <= 10000; i++) ve[i].clear();
memset(done, false, sizeof(done));
for(int i = 1; i < n; i++){
scanf("%d%d%d", &a, &b, &w);
ve[a].push_back(node(b, w));
ve[b].push_back(node(a, w));
}
ans = 0;
balance = inf;
numm = n;
getroot(1, 1);
//printf("root = %d\n", root);
dfs(root);
printf("%d\n", ans);
}
return 0;
}
