题目传送门

sol:看了题意显然是最大生成树，但是任意两个点之间都有边，大概有n*n条边。用朴素的最小生成树算法显然不行。联想了一下树的直径还是不会。看了大佬的题解，懂了。。。

• 树的直径
#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<int, LL> PIL;
const int MAXN = 1e5 + 10;
vector<PII> edge[MAXN];
LL _max; int nn1, nn2;
LL dis[MAXN];
PIL get_diameter(int rt, int fa) {
LL m1 = 0, m2 = 0;
int n1 = rt, n2 = rt;
for (PII i : edge[rt]) {
if (i.first == fa) continue;
PIL p = get_diameter(i.first, rt);
if (p.second + i.second > m1) {
m2 = m1, n2 = n1;
m1 = p.second + i.second;
n1 = p.first;
} else if (p.second + i.second > m2) {
m2 = p.second + i.second;
n2 = p.first;
}
}
if (m1 + m2 > _max) {
nn1 = n1, nn2 = n2;
_max = m1 + m2;
}
return {n1, m1};
}
void dfs(int rt, int fa, LL mm) {
dis[rt] = max(dis[rt], mm);
for (PII i : edge[rt]) {
if (i.first == fa) continue;
dfs(i.first, rt, mm + i.second);
}
}
int main() {
int n;
while (~scanf("%d", &n)) {
memset(dis, -1, sizeof(dis));
for (int i = 1; i <= n; i++) edge[i].clear();
for (int i = 1; i < n; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
edge[u].push_back({v, w});
edge[v].push_back({u, w});
}
_max = -1;
get_diameter(1, -1);    // 求出树的直径，以及两个端点；
dfs(nn1, -1, 0), dfs(nn2, -1, 0);
LL sum = 0;
for (int i = 1; i <= n; i++) sum += dis[i];
printf("%lld\n", sum - _max);    // 将两个端点加入集合只用算一次直径，而上面的循环算了两次，所以减掉一个直径；
}
return 0;
}

posted @ 2019-10-04 21:56  Angel_Demon  阅读(60)  评论(0编辑  收藏