Codeforces 1120D (树形DP 或 最小生成树)

题意看这篇博客：https://blog.csdn.net/dreaming__ldx/article/details/88418543

思路看这篇：https://blog.csdn.net/corsica6/article/details/88115948

有个坑点，不能深搜去找具体方案，不然 test14 会 MLE(或许是本蒟蒻写丑了)

代码：

#include <bits/stdc++.h>
#define LL long long
#define pii pair<int, int>
using namespace std;
const int maxn = 200010;
int head[maxn], Next[maxn * 2], ver[maxn * 2], tot;
int a[maxn];
LL dp[maxn][2], sum[maxn], f[maxn];
bool v[maxn][2];
bool res[maxn];
vector<int> ans;
bool is_leaf[maxn];
void add(int x, int y) {
	ver[++tot] = y;
	Next[tot] = head[x];
	head[x] = tot;
}
void dfs1(int x, int fa) {
	int cnt = 0;
	f[x] = fa;
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa) continue;
		dfs1(y, x);
		sum[x] += dp[y][0];
		cnt++;
	}
	if(cnt == 0) {
		dp[x][0] = a[x];
		dp[x][1] = 0;
		is_leaf[x] = 1;
		return;
	}
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa) continue;
		dp[x][0] = min(dp[x][0], sum[x] - dp[y][0] + dp[y][1] + a[x]);
		dp[x][1] = min(dp[x][1], sum[x] - dp[y][0] + dp[y][1]);
	}
	dp[x][0] = min(dp[x][0], sum[x]);
}

void bfs() {
	queue<pii> q;
	q.push(make_pair(1, 0));
	while(!q.empty()) {
		pii tmp = q.front();
		q.pop();
		int x = tmp.first, flag = tmp.second;
		if(v[x][flag]) continue;
		v[x][flag] = 1;
		if(flag == 0) {
			int pos = -1, cnt = 0;
			if(is_leaf[x]) {
				res[x] = 1;
				continue;
			}
			for (int i = head[x]; i; i = Next[i]) {
				int y = ver[i];
				if(y == f[x]) continue;
				if(dp[x][flag] == sum[x] - dp[y][0] + dp[y][1] + a[x]) {
					if(v[y][1]) continue;
					res[x] = 1;
					q.push(make_pair(y, 1));
					pos = y;
					cnt++;
				}
			}
			for (int i = head[x]; i; i = Next[i]) {
				int y = ver[i];
				if(v[y][0]) continue;
				if(y == f[x] || y == pos) continue;
				q.push(make_pair(y, 0));
			}
			if(cnt > 1 || (sum[x] == dp[x][0] && pos != -1)) {
				if(v[pos][0]) continue;
				q.push(make_pair(pos, 0));
			}
		} else {
			int pos = -1, cnt = 0;
			for (int i = head[x]; i; i = Next[i]) {
				int y = ver[i];
				if(y == f[x]) continue;
				if(dp[x][flag] == sum[x] - dp[y][0] + dp[y][1]) {
					if(v[y][1]) continue;
					q.push(make_pair(y, 1));
					pos = y;
					cnt++;
				}
			}
			for (int i = head[x]; i; i = Next[i]) {
				int y = ver[i];
				if(v[y][0]) continue;
				if(y == f[x] || y == pos) continue;
				q.push(make_pair(y, 0));
			}
			if(cnt > 1) {
				if(v[pos][0]) continue;
				q.push(make_pair(pos, 0));
			}
		}
	}
}
int main() {
	int n, x, y;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	memset(dp, 0x3f, sizeof(dp));
	dfs1(1, -1);
	bfs();
	for (int i = 1; i <= n; i++) {
		if(res[i])
			ans.push_back(i);
	}
	printf("%lld %d\n", dp[1][0], ans.size());
	sort(ans.begin(), ans.end());
	for (int i = 0; i < ans.size(); i++)
		printf("%d ", ans[i]);
	printf("\n");
}

最小生成树解法先留个坑在这。。。

补坑：

思路看这篇博客：https://www.cnblogs.com/river-flows-in-you/p/10596821.html

说一下我个人的理解：把每个叶子节点看成新图的顶点，对于原树中的每个顶点，我们可以计算出它影响哪些叶子节点，用差分的思想连边。只要求出了生成树就说明可以任意取，因为形成生成树之后，我们对每个点的赋值操作就可以类比在生成树上遍历的过程。

代码：

#include <bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 200010;
int head[maxn], Next[maxn * 2], ver[maxn * 2], tot;
int sz[maxn], cnt, l[maxn], r[maxn], a[maxn], f[maxn];
bool v[maxn];
struct Edge{
	int u, v, w, id;
	bool operator < (const Edge& rhs) const {
		return w < rhs.w;
	}
};
Edge b[maxn];
void add(int x, int y) {
	ver[++tot] = y;
	Next[tot] = head[x];
	head[x] = tot;
}
void dfs(int x, int fa) {
	sz[x] = 1;
	l[x] = INF, r[x] = -1;
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];
		if(y == fa) continue;
		dfs(y, x);
		sz[x] += sz[y];
		l[x] = min(l[x], l[y]);
		r[x] = max(r[x], r[y]);
	}
	if(sz[x] == 1) {
		l[x] = r[x] = ++cnt;
	}
	b[x] = (Edge){l[x], r[x] + 1, a[x], x};
}
int get(int x) {
	if(x == f[x]) return x;
	return f[x] = get(f[x]);
}
int main() {
	int n, x, y;
	scanf("%d", &n);
	int sum = 0;
	LL ans = 0;
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, -1);
	sort(b + 1, b + 1 + n);
	cnt++;
	for (int i = 1; i <= cnt; i++) f[i] = i;
	for(int L = 1, R; L <= n; L = R + 1) {
		R = L;
		while(b[L].w == b[R + 1].w && R < n) R++;
		for (int i = L; i <= R; i++) {
			x = get(b[i].u), y = get(b[i].v);
			if(x != y) {
				v[b[i].id] = 1;
				sum++;
			}
		}
		for (int i = L; i <= R; i++) {
			x = get(b[i].u), y = get(b[i].v);
			if(x != y) {
				ans += b[i].w;
//				printf("%lld %d\n", ans, b[i].w);
				f[x] = y;
			}
		}
	}
//	printf("%lld %d\n", ans, sum);
	cout << ans << " " << sum << endl;
	for (int i = 1; i <= n; i++)
		if(v[i]) printf("%d ", i);
}