【bzoj1468】Tree

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*题目描述：
给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K
*输入：
N（n<=40000）接下来n-1行边描述管道，按照题目中写的输入接下来是k
*输出：
一行，有多少对点之间的距离小于等于k
*样例输入：
7
1 6 13
6 3 9
3 5 7
4 1 3
2 4 20
4 7 2
10
*样例输出：
5
*题解：
点分治。就是把树上的路径统计问题分成过某个顶点和不过这个点的来做。不过这个点的就递归下去做，过这个点的就直接做。每次递归的时候找到这棵树的重心然后继续往下做，这样时间复杂度就有保证了，每一次找到重心后，这棵树的size至少缩小了一半，于是乎点分治的时间复杂度则降为了O(nlog2n)。不过这题统计过一个点的路径时需要排序（貌似可以用基排？反正我懒得手写），于是乎这题的复杂度就为O(nlog22n)的了（如果是基数排序的话就为O(nlog2n)）。
*代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#ifdef WIN32    #define LL "%I64d"#else    #define LL "%lld"#endif#ifdef CT    #define debug(...) printf(__VA_ARGS__)    #define setfile() #else    #define debug(...)    #define filename ""    #define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);#endif#define R register#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)char B[1 << 15], *S = B, *T = B;inline int FastIn(){    R char ch; R int cnt = 0; R bool minus = 0;    while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;    ch == '-' ? minus = 1 : cnt = ch - '0';    while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';    return minus ? -cnt : cnt;}#define maxn 300010#define maxm 600010struct Edge{    Edge *next;    int to, w;}*last[maxn], e[maxm], *ecnt = e;inline void link(R int a, R int b, R int v){    *++ecnt = (Edge) {last[a], b, v}; last[a] = ecnt;}int n, k, size[maxn], son[maxn], root, sum, depcnt;long long ans, deep[maxm], d[maxn];bool vis[maxn];void dfsroot(R int x, R int fa){    size[x] = 1; son[x] = 0;    for (R Edge *iter = last[x]; iter; iter = iter -> next)    {        R int pre = iter -> to;        if (pre == fa || vis[pre]) continue;        dfsroot(pre, x);        size[x] += size[pre];        cmax(son[x], size[pre]);    }    cmax(son[x], sum - size[x]);    if (root == 0 || son[x] < son[root]) root = x;}void dfsdep(R int x, R int fa){    deep[++depcnt] = d[x];    for (R Edge *iter = last[x]; iter; iter = iter -> next)    {        R int pre = iter -> to;        if (pre == fa || vis[pre]) continue;        d[pre] = d[x] + iter -> w;        dfsdep(pre, x);    }}inline long long cal(R int x, R int val){    d[x] = val; depcnt = 0;    dfsdep(x, 0);    std::sort(deep + 1, deep + depcnt + 1);    R long long t = 0;    R int l, r;    for (l = 1, r = depcnt; l < r; )    {        if (deep[l] + deep[r] <= k) {t += r - l; ++l;}        else --r;    }    return t;}void work(R int x){    ans += cal(x, 0);    vis[x] = 1;    for (R Edge *iter = last[x]; iter; iter = iter -> next)    {        R int pre = iter -> to;        if (vis[pre]) continue;        ans -= cal(pre, iter -> w);        sum = size[pre];        root = 0;        dfsroot(pre, 0);        work(root);    }}int main(){    n = FastIn(); sum = n;    for (R int i = 1; i < n; ++i)    {        R int a = FastIn(), b = FastIn(), v = FastIn();        link(a, b, v); link(b, a, v);    }    k = FastIn();    son[root = 0] = 1 << 30;    dfsroot(1, 0);    work(root);    printf("%lld\n", ans );    return 0;}

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