POJ - 1741 Tree 树的分治

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题目大意：给出一棵N个点的树，每条边都有相应的权值。
先给出K，要求你找出权值小于等于k的（u,v）对

解题思路：具体的思路可以参考漆子超的《分治算法在树的路径问题中的应用》这篇论文。

#include<cstdio>#include<algorithm>#include<vector>using namespace std;#define maxn 10010vector<int> Node[maxn], W[maxn], dep;int Sum[maxn], dp[maxn], vis[maxn], d[maxn];int n, k, root, size, Ans;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]] && Node[cur][i] != fa) {            dfs(Node[cur][i], cur);            Sum[cur] += Sum[Node[cur][i]];            dp[cur] = max(dp[cur], Sum[Node[cur][i]]);        }    }    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();        W[i].clear();        vis[i] = 0;    }    int x, y, z;    for(int i = 0; i < n - 1; i++) {        scanf("%d%d%d", &x, &y, &z);        Node[x].push_back(y);        Node[y].push_back(x);        W[x].push_back(z);        W[y].push_back(z);    }    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(Node[cur][i] != fa && !vis[Node[cur][i]]) {            d[Node[cur][i]] = d[cur] + W[cur][i];            getdep(Node[cur][i], 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]]) {            Ans -= calc(Node[cur][i], W[cur][i]);            dp[0] = size = Sum[Node[cur][i]];            dfs(Node[cur][i], root = 0);            solve(root);        }    }}int main() {    while(scanf("%d%d", &n, &k) != EOF && n + k) {        init();        dfs(1,0);        solve(root);        printf("%d\n", Ans);    }    return 0;}

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