POJ - 1987 Distance Statistics 树上的分治

题目大意：和poj 1741的那题和类似，求树上节点之间的距离小于等于k的节点对有多少对

解题思路：具体可参考：《分治算法在树的路径问题中的应用——漆子超》
给这题的输入坑了，注意输入，不然会超时

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 40010
int vis[maxn], Sum[maxn], d[maxn], dp[maxn];
int n, k, size, root, m, Ans;
struct node{
    int v, l;
    node() {}
    node(int vv, int ll) {
        v = vv;
        l = ll;
    }
};
vector<node> Node[maxn]; 
vector<int> dep;

void dfs(int cur, int fa) {
    Sum[cur] = 1;
    dp[cur] = 0;
    for(int i = 0; i < Node[cur].size(); i++) {
        if(!vis[Node[cur][i].v] && Node[cur][i].v != fa) {
            dfs(Node[cur][i].v, cur);
            Sum[cur] += Sum[Node[cur][i].v];
            dp[cur] = max(dp[cur], Sum[Node[cur][i].v]);
        }
    }
    dp[cur] = max(dp[cur], size - Sum[cur]);
    if(dp[cur] < dp[root])
        root = cur;
}

void init() {

    for(int i = 1; i <= n; i++) {
        Node[i].clear();
        vis[i] = 0;
    }

    int x, y, z;
    char c;
    for(int i = 0; i < m; i++) {
        scanf("%d%d%d %c", &x, &y, &z, &c);
        Node[x].push_back(node(y,z));
        Node[y].push_back(node(x,z));
    }
    scanf("%d", &k);
    dp[0] = size = n;
    Ans = root = 0;
}

void getdep(int cur, int fa) {
    dep.push_back(d[cur]);
    for(int i = 0; i < Node[cur].size(); i++)
        if(!vis[Node[cur][i].v] && Node[cur][i].v != fa) {
            d[Node[cur][i].v] = d[cur] + Node[cur][i].l;
            getdep(Node[cur][i].v, cur);
        }
}

int calc(int cur, int Num) {

    dep.clear();
    d[cur] = Num;
    getdep(cur, 0);
    sort(dep.begin(), dep.end());
    int cnt = 0;
    for(int l = 0, r = dep.size() - 1; l < r; )
        if(dep[l] + dep[r] <= k)
            cnt += r - l++;
        else
            r--;
    return cnt;
}

void solve(int cur) {

    Ans += calc(cur, 0);
    vis[cur] = 1;
    for(int i = 0; i < Node[cur].size(); i++) {
        if(!vis[Node[cur][i].v]) {
            Ans -= calc(Node[cur][i].v, Node[cur][i].l);
            dp[0] = size = Sum[Node[cur][i].v];
            root = 0;
            dfs(Node[cur][i].v, 0);
            solve(root);
        }
    }
}

int main() {
    scanf("%d%d", &n, &m);
    init();
    dfs(1,0);
    solve(root);
    printf("%d\n", Ans);
    return 0;
}