POJ

传送门
//就是求树上距离小于等于k的点对有多少对.
//就是树的点分治模板题, 推荐ioi国家集训队论文, 里面讲解的非常清楚了. 我就不多说了. 存个板子.

//树的点分治，该代码是求树上有多少对点的dis<=k//vis代表该点是否已经当做过重心，siz是子树节点个数、mv是子树中最大的节点数int n, cnt, head[maxn], k, vis[maxn], root, maxx, dis[maxn];int ans, num, tot, siz[maxn], mv[maxn]; //tot代表当前树的节点个数、如果遇到时间过长，考虑重心是否找对struct node {    int to, w, next;} e[maxn<<1];void add(int u, int v, int w){    e[cnt] = (node){v,w,head[u]};    head[u] = cnt++;}void getroot(int u, int fa){    siz[u] = 1, mv[u] = 0;    for (int i = head[u]; ~i; i = e[i].next) {        int to = e[i].to;        if (to == fa || vis[to]) continue;        getroot(to, u);        siz[u] += siz[to];        mv[u] = max(mv[u], siz[to]);    }    mv[u] = max(mv[u], tot - siz[u]);    if (mv[u] < mv[root]) root = u;}void getdis(int u,int fa,int dep){    dis[++num] = dep;    for (int i = head[u]; ~i; i = e[i].next) {        int to = e[i].to;        if (to == fa || vis[to]) continue;        getdis(to, u, dep + e[i].w);    }}int cal(int u,int f){    int res = 0;    num = 0;    getdis(u,-1,f);    sort(dis+1,dis+num+1);    /*if(u == 1) {        cout << "###" << num << endl;        for(int i=1;i<=num;i++){            printf("%d%c",dis[i],i==num?'\n':' ' );        }    }*/    int r = num ;    for(int l = 1; l < r; l++) {        while(dis[l] + dis[r] > k && l < r)            r--;        res += r - l ;    }    return res;}void work(int u){    vis[u] = 1;    ans += cal(u, 0);    //if(u == 1) cout << "!!!" << ans << endl;    for (int i = head[u]; ~i; i = e[i].next) {        int to = e[i].to;        if (vis[to]) continue;        int tmp = cal(to, e[i].w);        //传e[i].w是因为这是根节点到该点的距离,因为要和k判断关系,所以必须加上        //if(u == 1) cout << "@@@" << tmp << endl;        ans -= tmp;        mv[root=0] = tot = siz[to];        getroot(to, -1);        work(root);    }}void solve(){    while(~scanf("%d%d",&n,&k)){        if (n + k == 0 ) break;        cnt =0 ; Fill(head,-1); Fill(vis,0);        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);        }        ans = 0 ;        mv[root=0] = tot = n;        getroot(1, -1);        work(root);        cout << ans << endl;    }}