SPOJ Two Paths

Description

给定一个无向图，含有一定的路。从中找出两个最长的路径（每条路径有一些相通路组成）这两个路径不能经过公共的点，求何时二路径的乘积最大。

本题给出的无向图是一棵树，每边权值为1.

原文翻译应为有n个点，n-1条边，两点之间能够相互到达。

Solution

• 直径分成两部分得到的两条路径
• 直径的一部分和另一部分里的最长链

Code

#include <vector>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const int N = 100005;

namespace { // 此处应该替换为Class Solution
    int n, dep[N], fa[N], MxDepP;
    int dis[N], cnt, d[N], ind[N];
    int L[N], R[N];

    std:: vector<int> e[N];

    void dfs1(int u, int fath) {
        dep[u] = dep[fath] + 1, fa[u] = fath;
        for (auto v : e[u])
            if (v != fath)
                dfs1(v, u);
        if (dep[u] > dep[MxDepP])
            MxDepP = u;
    }

    void dfs2(int u, int fath) {
        for (int v : e[u])
            if (v != fath and not ind[v])
                dfs2(v, u), dis[u] = std:: max(dis[u], dis[v] + 1);
    }
    int GetDiameter() {
        dfs1(1, 0);
        dfs1(MxDepP, 0);
        int end = MxDepP;
        while(end)
            d[++cnt] = end, ind[end] = true, end = fa[end];
    }
    long long CalcAnswer() {
        long long Res = 0;
        for (int i = 1; i <= cnt; i += 1) {
            dfs2(d[i], 0);
            Res = std:: max(Res, 1ll * (dis[d[i]] - 1) * (cnt - 1));
        }
        int tmp = 0;
        for (int i = 1; i <= cnt; i += 1)
            L[i] = tmp = std:: max(tmp, i - 2 + dis[d[i - 1]]);
        tmp = 0;
        for (int i = cnt; i >= 1; i -= 1)
            R[i] = tmp = std:: max(tmp, cnt - i + dis[d[i + 1]] - 1);
        for (int i = 1; i <= n; i += 1)
            Res = std:: max(Res, 1ll * L[i] * R[i - 1]);
        return Res;
    }
};

int main () {
    scanf("%d", &n);
    for (int i = 1; i < n; i += 1) {
        int u, v;
        scanf("%d%d", &u, &v);
        e[u].push_back(v), e[v].push_back(u);
    }
    GetDiameter();
    printf("%lld\n", CalcAnswer());
    return 0;
}