SPOJ Two Paths(树形dp，最大不相交路径长度乘积)

题目链接：
SPOJ Two Paths
题意：
给一个n个节点和n−1条边的树，求两条不相交（无公共节点）的路径长度乘积最大值？（路径长度就是路径上边的数量）
数据范围：n≤105
分析：
这道题和Codeforces 633 F The Chocolate Spree是其实一样的。本来以为会好些点，实际上还是写了好久。。。。主要是细节太多了，有的地方数组的定义也不大一样，不多说了。。。

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <queue>#include <vector>using namespace std;typedef long long ll;const int MAX_N = 100010;int n, total;int head[MAX_N];ll up[MAX_N], down[MAX_N], best[MAX_N];ll predown[MAX_N], ppredown[MAX_N], sufdown[MAX_N], ssufdown[MAX_N];ll prebest[MAX_N], sufbest[MAX_N];vector<int> child;struct Edge {    int to, next;}edge[MAX_N * 2];inline void AddEdge(int from, int to){    edge[total].to = to;    edge[total].next = head[from];    head[from] = total++;}//down[i]:从i到叶子节点的最长路径,ddown:次长路径//best[i]:i子树内部的最长链,best可以经过根节点void dfs(int u, int p){    ll Max = 0, MMax = 0;    int cnt = 0;    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (v == p) continue;        dfs(v, u);        cnt++;        if (down[v] > Max) {            MMax = Max;            Max = down[v];        } else if (down[v] > MMax) {            MMax = down[v];        }        best[u] = max(best[u], best[v]);    }    if (cnt == 0) return;    down[u] = Max + 1;    if (cnt > 1) best[u] = max(best[u], Max + MMax + 2); //细节    else best[u] = max(best[u], Max + MMax + 1);  //细节    //printf("down[%d] = %lld best[%d] = %lld\n", u, down[u], u, best[u]);}void solve(){    ll ans = 0;    queue<pair<int, int> > que;    que.push(make_pair(1, -1));    while (!que.empty()) {        pair<int, int> cur = que.front();        que.pop();        int u = cur.first, p = cur.second;        child.clear();        child.push_back(0);        for (int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].to;            if (v == p) continue;            child.push_back(v);        }        int size = child.size();        // 前缀down最大和次大,前缀best最大        // predown和ppredown包括了兄弟节点到父亲的路径        prebest[0] = predown[0] = ppredown[0] = 0;        for (int i = 1; i < size; ++i) {            int v = child[i];            prebest[i] = max(prebest[i - 1], best[v]);            predown[i] = predown[i - 1], ppredown[i] = ppredown[i - 1];            if (down[v] + 1 > predown[i]) {                ppredown[i] = predown[i];                predown[i] = down[v] + 1;            } else if (down[v]+ 1 > ppredown[i]) {                ppredown[i] = down[v] + 1;            }        }        sufdown[size] = ssufdown[size] = sufbest[size] = 0;        for (int i = size - 1; i >= 1; --i) {            int v = child[i];            sufbest[i] = max(sufbest[i + 1], best[v]);            sufdown[i] = sufdown[i + 1], ssufdown[i] = ssufdown[i + 1];            if (down[v] + 1 > sufdown[i]) {                ssufdown[i] = sufdown[i];                sufdown[i] = down[v] + 1;            } else if (down[v] + 1 > ssufdown[i]) {                ssufdown[i] = down[v] + 1;            }        }        //up[i]包含i到i的父亲的路径        for (int i = 1; i < size; ++i) {            int v = child[i];            ll outside = up[u] + max(predown[i - 1], sufdown[i + 1]);            outside = max(outside, predown[i - 1] + ppredown[i - 1]);            outside = max(outside, sufdown[i + 1] + ssufdown[i + 1]);            outside = max(outside, predown[i - 1] + sufdown[i + 1]);            outside = max(outside, prebest[i - 1]);            outside = max(outside, sufbest[i + 1]);            //printf("v = %d outside = %lld best[v] = %lld\n", v, outside, best[v]);            ans = max(ans, outside * best[v]);        }        // predown/sufdown 算上了兄弟节点到父亲节点的路径        // up[v]包含了v到父亲u的路径        for (int i = 1; i < size; ++i) {            int v = child[i];            up[v] = 1 + max(up[u], max(predown[i - 1], sufdown[i + 1]));            que.push(make_pair(v, u));        }    }    printf("%lld\n", ans); }int main(){    freopen("G.in", "r", stdin);    while (~scanf("%d", &n)) {        memset(head, -1, sizeof(head));        total = 0;        for (int i = 1; i < n; ++i) {            int u, v;            scanf("%d%d", &u, &v);            AddEdge(u, v);            AddEdge(v, u);        }        memset(down, 0, sizeof(down));        memset(best, 0, sizeof(best));        memset(up, 0, sizeof(up));        dfs(1, -1);        solve();    }    return 0;}