fzu 2038 Another Postman Problem(递归求解)

题目链接：fzu 2038 Another Postman Problem

题目大意：给出n， 然后给出n - 1条边，保证图联通，计算每个点到其他所有点的路径总和的和。

解题思路：一开始想到Floyd算法求出所有点点之间的最短路，但是o(n^3)的复杂度太高了;后来想到说用枚举起点后直接用BFS去计算距离，可是o（n^2)还是超时。后来灵光一闪，发现可以枚举每条边走过的次数，以为图肯定为无环图，所以剪断当前边后，可以将图上的点分为两个集合，然后走过的次数就是两个集合个数的乘积。然后递归遍历每个点的每条边。

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>

using namespace std;

const int N = 100005;
typedef long long ll;

struct node {
int next;
ll val;
ll size;
node() { next = 0, val = size = 0; };
node(int next, ll val) {
this->next = next;
this->val = val;
this->size = 0;
}
};

int vis
;
ll n;
vector<node> v
;

void init() {
scanf("%lld", &n);
for (int i = 0; i < n; i++) v[i].clear();
memset(vis, 0, sizeof(vis));

int a, b;
ll c;
for (int i = 1; i < n; i++) {
scanf("%d%d%lld", &a, &b, &c);
v[a].push_back(node(b, c));
v[b].push_back(node(a, c));
}
}

ll dfs(int s) {
ll cnt = 0;

vis[s] = 1;
for (int i = 0; i < v[s].size(); i++) {
int u = v[s][i].next;
if (vis[u]) continue;

ll t = dfs(u);
v[s][i].size = (n - t) * t;
cnt += t;
}

return cnt + 1;
}

ll solve() {
dfs(0);
ll ans = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < v[i].size(); j++)
ans += v[i][j].val * v[i][j].size;
}
return ans * 2;
}

int main () {
int cas;
scanf("%d", &cas);
for (int i = 1; i <= cas; i++) {
init();
printf("Case %d: %lld\n", i, solve());
}
return 0;
}