21.10.17模拟 绵羊

给出一个完全图唯一最小生成树，求完全图所有边权的和
考场想到n^2log的计算，枚举两点再求出两点路上最长边再判断要不要+1.没想到就是个kruskal水题

const int N = 23007;
struct edge {
	int x, y, z;
	bool operator <(const edge&rhs) const {
		return z < rhs.z;
	}
} t[N];
int n, T;
int fa[N], siz[N];

inline int find(int x) {
	while(x != fa[x]) x = fa[x] = fa[fa[x]];
	return x;
}

int main() {
	freopen("sheep.in","r",stdin);
	freopen("sheep.out","w",stdout);
	read(T);
	while(T--) {
		read(n);
		rep(i, 1, n) {
			fa[i] = i;
			siz[i] = 1;
		}
		lxl ans(0);
		rep(i, 2, n) {
			read(t[i - 1].x);
			read(t[i - 1].y);
			read(t[i - 1].z);
			ans += t[i - 1].z;
		}
		std::sort(t + 1, t + n);
		rep(i, 1, n - 1) {
			int x(find(t[i].x));
			int y(find(t[i].y));
			if(x == y) continue;
			ans += 1ll * (t[i].z + 1) * (siz[x] * siz[y] - 1);
			fa[x] = y;
			siz[y] += siz[x];
		}
		out(ans, '\n');
	}
	return 0;
}

标签：2log,const,17,水题,int,绵羊,21.10,两点,operator
来源： https://www.cnblogs.com/QQ2519/p/15419240.html