Holiday Accommodation SPOJ - HOLI (center of gravity tree / right side contribution)

Holiday Accommodation

Meaning of the questions: a weighted tree, so that each node in the tree are by the shortest walk to another location, and all the nodes and the maximum path distance traveled. And seeking the maximum distance.

Problem solution 1: Find the center of gravity of the tree, and the answer is all of the nodes to the center of gravity and the distance twice.

Problem solution 2: Consider an edge contribution to the answer, is the number of points at both ends of the right side edge lower value * * 2. Enumerate all sides to calculate the contribution. Code for using this method.

#include <bits/stdc++.h>
#define fopi freopen("in.txt", "r", stdin)
#define fopo freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 100;
typedef pair<int, int> Pair;

vector<Pair> V[maxn];
LL sum[maxn], ans;

void dfs1(int x, int from) {
    sum[x] = 1;
    for (auto p : V[x]) {
        int y = p.first;
        if (y == from) continue;
        dfs1(y, x);
        sum[x] += sum[y];
    }
}

void dfs2(int x, int from) {
    for (auto p : V[x]) {
        int y = p.first, l = p.second;
        if (y == from) continue;
        ans += min(sum[y], sum[1] - sum[y]) * l * 2;
        dfs2(y, x);
    }
}

int T, n;
int main() {
    //fopi;

    scanf("%d", &T);
    for (int ca = 1; ca <= T; ca++) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) V[i].clear();

        for (int i = 1; i <= n-1; i++) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            V[x].push_back({y, z});
            V[y].push_back({x, z});
        }

        ans = 0;
        dfs1(1, 0);
        dfs2(1, 0);

        printf("Case #%d: %lld\n", ca, ans);
    }
}