【BZOJ1468】Tree

Tree

Description

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

Input

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

Output

一行，有多少对点之间的距离小于等于k

Sample Input

7

1 6 13

6 3 9

3 5 7

4 1 3

2 4 20

4 7 2

10

Sample Output

5

Source

LTC男人八题系列（POJ1741）
TMD为什么八中能A,POJ就是T！T！T！

Solution

点分治 对于一条树路径 只有经过或不经过一个点的情况
对于不经过的情况 把一棵树按这个点拆成好几棵分治就行了
考虑经过这个点的情况
对于这题 可以对这个点延伸出的几棵子树各做一次dfs
记录子树中出现的距离值
对于一棵树的距离值数组
把它排序求一次ans1
再对每棵子树分别求一个自己对自己的ans2
ans1-Σans2即为最后的ans

Code

/**************************************************************
    Problem: 1468
    User: johann
    Language: C++
    Result: Accepted
    Time:876 ms
    Memory:4284 kb
****************************************************************/

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define red(i, a, b) for(int i = (a); i >= (b); i--)
#define ll long long

inline int read() {
    int x = 0; char c = getchar();
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return x;
}

const int N = 55000;
const int inf = 1000000000;
struct edge{
    int from, to, len, nxt;
}e[N * 2];
int depth[N], head[N], size[N], mx[N], d[N], vis[N];
int n, m, k, tail = 0, sum, ans, root;

void addedge(int x, int y, int z) {
    e[++tail].from = x;
    e[tail].to = y;
    e[tail].len = z;
    e[tail].nxt = head[x];
    head[x] = tail;
}

void getroot(int u, int fa) {
    size[u] = 1; mx[u] = 0;
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if (v == fa || vis[v]) continue;
        getroot(v, u);
        size[u] += size[v];
        mx[u] = max(mx[u], size[v]);
    }
    mx[u] = max(mx[u], sum - size[u]);
    if (mx[u] < mx[root]) root = u;
}

void getdep(int u, int fa) {
    depth[++m] = d[u];
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if (v == fa || vis[v]) continue;
        d[v] = d[u] + e[i].len;
        getdep(v, u);
    }
}

int calc(int u, int len) {
    d[u] = len; m = 0;
    getdep(u, 0);
    sort(depth + 1, depth + m + 1);
    int ret = 0, l, r;
    for(l = 1, r = m; l < r;) {
        if (depth[l] + depth[r] <= k) {
            ret += r - l;
            l++;
        }else r--;
    }
    return ret; 
}

void work(int u) {
    ans += calc(u, 0);
    vis[u] = 1;
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if (vis[v]) continue;
        ans -= calc(v, e[i].len);
        sum = size[v];
        root = 0;
        getroot(v, 0);
        work(v);
    }
}

int main() {
    n = read();
    tail = root = 0; ans = 0;
    memset(vis, 0, sizeof(vis));
    rep(i, 1, n) head[i] = -1;
    rep(i, 1, n - 1) {
        int x = read(), y = read(), z = read();
        addedge(x, y, z);
        addedge(y, x, z);
    }
    k = read();
    sum = n;
    mx[0] = inf;
    getroot(1, 0);
    work(root);
    printf("%d\n", ans);
    return 0;
}

End.