man turn str truct case 分治 lines ssi and

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
題目分析 ： 給定一棵樹，以及樹上邊的關系大小，問你又多少對點的距離是小於等於所給定的 d 的
思路分析 ： 樹上點分治的板子題，首先尋找樹的重心，以重心為根結點，尋求所有符合題意要求的點對，但是這樣計算會算出一些不符合題目的點對，在減去即可，此時當遍歷到一個新的結點時，此時的情況又可以當成最初的情況，找重心的時候要註意，對它的子樹來說，總的結點數是小於 n 的！！！
代碼示例 ：

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表示每個結點所連的結點數， mx表示對每個根結點所連的最大結點子樹有多少的結點
int balance;
bool done[maxn];
int ans = 0;
int numm; // 表示結點總數

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]); // 對子樹在尋找子樹的重心的過程中，子樹的總結點數是會變小的
    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]; // 這裏是重點，因為這個地方一直T，還以為寫的代碼有問題
        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;
}

樹上點分治 poj 1741