HDU 5242 树链剖分思想的贪心

题意及博客

树链剖分分为2步，第一次求出深度，重儿子，第二次求出重链，用到了启发式的思想，即对于比较重的儿子，尽量去完整的维护它。类似于我们去合并两个堆，明显把小的堆逐个插入大的堆中会比大的往小的插更优，而这可以达到均摊O(logn)的效果。对于这个题，类似选重儿子， 我们每次尽量选择最长的路径，选出前m个就可以了。

代码：

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 100010;
int head[maxn], Next[maxn * 2], ver[maxn * 2], tot;
LL mx_d[maxn], w[maxn], dist[maxn], d[maxn];
void add(int x, int y) {
	ver[++tot] = y;
	Next[tot] = head[x];
	head[x] = tot;
}

void dfs1(int x, int fa = 0) {
	mx_d[x] = d[x];
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y != fa) {
			d[y] = d[x] + w[y];
			dfs1(y, x);
			mx_d[x] = max(mx_d[x], mx_d[y]);
		}
	}
}
void dfs2(int x, int fa, LL now_dist) {
	LL mx = -1, tmp = 0;
	dist[x] = now_dist;
	int pos = 0;
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa) continue;
		if(mx_d[x] == mx_d[y]) {
			pos = y;
			break;
		}
	}
	if(pos != 0) {
		dfs2(pos, x, dist[x] + w[pos]);
		dist[x] = 0;
	}
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa || y == pos) continue;
		dfs2(y, x, w[y]);
	}
}
int main() {
	int T, n, m, x, y, kase = 0;
	cin >> T;
	while(T--) {
		memset(head, 0, sizeof(head));
		tot = 0;
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) {
			scanf("%lld", &w[i]);
		}
		d[1] = dist[1] = w[1];
		for (int i = 1; i < n; i++) {
			scanf("%d%d", &x, &y);
			add(x, y);
			add(y, x);
		}
		dfs1(1);
		dfs2(1, 0, w[1]);
		LL ans = 0;
		sort(dist + 1, dist + n + 1);
		for (int i = n; i >= n - m + 1; i--) {
			ans += dist[i];
		}
		printf("Case #%d: ", ++kase);
		printf("%lld
", ans);
	}
	
}