T20201119

Posted on Edited on In Word count in article: 1.8k Reading time ≈ 7 mins.
大东辰的测试

T1

一来就是一个换根 \(dp\)，然后很难调，细节容易挂。。。

还是先考虑以 \(1\) 为根的答案，然后需要考虑它的子树该如何遍历能使最后的答案更优。

假设两颗子树 \(a,b\)，他们子树内的答案、大小、权值之和分别表示为 \(f,siz,sum\)，\(a\) 在 \(b\) 前更优当且仅当 \(sum_b\times siz_a<sum_a\times siz_b\) 成立时成立。

所以，这一个 \(\text{dfs}\) 需要对子节点排一次序才能计算答案，即：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
void dfs(int u, int fa)
{
	siz[u] = 1, f[u] = sum[u] = a[u];
	for (int i = head[u]; i; i = nxt[i]) {
		if (edge[i] == fa) continue;
		dfs(edge[i], u);
		tmp[u].push_back(edge[i]);
	}

	sort (tmp[u].begin(), tmp[u].end(), cmp);
	for (int i = 0; i < tmp[u].size(); ++i) {
		f[u] += f[tmp[u][i]] + siz[u] * sum[tmp[u][i]];
		siz[u] += siz[tmp[u][i]], sum[u] += sum[tmp[u][i]];
	}
}

那么换根的时候，原来父亲节点的信息也需要相应的发生变化：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
void chroot(int u, int fa)
{
	sort (tmp[u].begin(), tmp[u].end(), cmp);
	siz[u] = 1, g[u] = sum[u] = a[u];
	for (int i = 0; i < tmp[u].size(); ++i) {
		g[u] += f[tmp[u][i]] + siz[u] * sum[tmp[u][i]];
		siz[u] += siz[tmp[u][i]], sum[u] += sum[tmp[u][i]];
	}
	ans = min(ans, g[u]);

	ll tempa(a[u]), temps(1);
	for (int i = 0; i < tmp[u].size(); ++i) {
		if (fa == tmp[u][i]) {
			tempa += sum[tmp[u][i]];
			temps += siz[tmp[u][i]];
			continue;
		}
		f[u] = g[u] - f[tmp[u][i]] - temps * sum[tmp[u][i]];
		tempa += sum[tmp[u][i]];
		temps += siz[tmp[u][i]];
		f[u] -= siz[tmp[u][i]] * (SUM - tempa);
		siz[u] = n - siz[tmp[u][i]];
		sum[u] = SUM - sum[tmp[u][i]];
		tmp[tmp[u][i]].push_back(u);
		chroot(tmp[u][i], u);
	}
}

所以这个点就可以直接当做根来算了。

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int maxn = 2e5 + 10;
int n, cur, cnt, a[maxn];
ll SUM, ans(0x7fffffffffffffff);
ll siz[maxn], sum[maxn], f[maxn], g[maxn];
int head[maxn], edge[maxn << 1], nxt[maxn << 1];
vector <int> tmp[maxn];

inline void addedge(int u, int v)
{
	nxt[++cur] = head[u];
	head[u] = cur;
	edge[cur] = v;
}

inline bool cmp(int a, int b)
{
	return siz[a] * sum[b] < siz[b] * sum[a];
}

void dfs(int u, int fa)
{
	siz[u] = 1, f[u] = sum[u] = a[u];
	for (int i = head[u]; i; i = nxt[i]) {
		if (edge[i] == fa) continue;
		dfs(edge[i], u);
		tmp[u].push_back(edge[i]);
	}

	sort (tmp[u].begin(), tmp[u].end(), cmp);
	for (int i = 0; i < tmp[u].size(); ++i) {
		f[u] += f[tmp[u][i]] + siz[u] * sum[tmp[u][i]];
		siz[u] += siz[tmp[u][i]], sum[u] += sum[tmp[u][i]];
	}
}

void chroot(int u, int fa)
{
	sort (tmp[u].begin(), tmp[u].end(), cmp);
	siz[u] = 1, g[u] = sum[u] = a[u];
	for (int i = 0; i < tmp[u].size(); ++i) {
		g[u] += f[tmp[u][i]] + siz[u] * sum[tmp[u][i]];
		siz[u] += siz[tmp[u][i]], sum[u] += sum[tmp[u][i]];
	}
	ans = min(ans, g[u]);

	ll tempa(a[u]), temps(1);
	for (int i = 0; i < tmp[u].size(); ++i) {
		if (fa == tmp[u][i]) {
			tempa += sum[tmp[u][i]];
			temps += siz[tmp[u][i]];
			continue;
		}
		f[u] = g[u] - f[tmp[u][i]] - temps * sum[tmp[u][i]];
		tempa += sum[tmp[u][i]];
		temps += siz[tmp[u][i]];
		f[u] -= siz[tmp[u][i]] * (SUM - tempa);
		siz[u] = n - siz[tmp[u][i]];
		sum[u] = SUM - sum[tmp[u][i]];
		tmp[tmp[u][i]].push_back(u);
		chroot(tmp[u][i], u);
	}
}

inline int read()
{
	int x(0);
	char o(getchar());
	while (o < '0' || o  > '9') o = getchar();
	for (; o >= '0' && o <= '9'; o = getchar())
		x = (x << 1) + (x << 3) + (o ^ 48);
	return x;
}

int main()
{
	freopen ("signin.in", "r", stdin);
	freopen ("signin.out", "w", stdout);
	n = read();

	for (int i = 1, u, v; i < n; ++i) {
		u = read(), v = read();
		addedge (u, v), addedge (v, u);
	}
	
	for (int i = 1; i <= n; ++i)
		SUM += (a[i] = read());

	dfs(1, 0), chroot(1, 0);

	printf ("%lld\n", ans);
}

T2

就是问这一条边若要保证不改变其最小生成树的大小所需要满足的条件。

那么很显然，可以先跑一次最小生成树：

  • 那么对于不在生成树上的边，可以查询 \(u\rightarrow v\) 路径上的树边最大的权值。
  • 对于在生成树上的边，可以查询 \(u\rightarrow v\) 路径上非树边的最小权值。

证明的话，就是可以把查询到的那条边拿走，然后把这条边换上去，并不影响生成树的连通性以及大小。

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
#include <bits/stdc++.h>
#define ls (x << 1)
#define rs (x << 1 | 1)

using namespace std;

const int maxn = 1e5 + 10;
const int maxm = 1e6 + 10;
const int inf = 0x7f7f7f7f;
struct Node
{
	int u, v, w, p;
	bool operator < (const Node &t) const {
		return w < t.w;
	}
} Lines[maxm];
bool vis[maxm];
int n, m, cnt, cur, ans[maxm];
int head[maxn], edge[maxn << 1], cost[maxn << 1], nxt[maxn << 1];
int son[maxn], size[maxn], top[maxn], dep[maxn], f[maxn], pkg[maxn];
int mx[maxn << 2], mn[maxn << 2], tag[maxn << 2], id[maxn], len[maxn];

inline int mmin(int a, int b)
{
	return a < b ? a : b;
}

inline int mmax(int a, int b)
{
	return a > b ? a : b;
}

inline void mswap(int &a, int &b)
{
	a ^= b ^= a ^= b;
}

inline int read()
{
	int x(0);
	char o(getchar());
	while (o < '0' || o > '9') o = getchar();
	for (; o >= '0' && o <= '9'; o = getchar())
		x = (ls) + (x << 3) + (o ^ 48);
	return x;
}

inline int find(int x)
{
	if (f[x] == x) return x;
	return f[x] = find(f[x]);
}

void dfs(int u, int fa)
{
	f[u] = fa, size[u] = 1, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = nxt[i]) {
		if (edge[i] == fa) continue;
		dfs (edge[i], u);
		size[u] += size[edge[i]];
		if (size[edge[i]] > size[son[u]]) son[u] = edge[i], pkg[u] = cost[i];
	}
}

void dfs(int u, int tp, int lenth)
{
	top[u] = tp, id[u] = ++cnt, len[cnt] = lenth;
	if (son[u]) dfs(son[u], tp, pkg[u]);
	for (int i = head[u]; i; i = nxt[i]) {
		if (edge[i] == f[u] || edge[i] == son[u]) continue;
		dfs(edge[i], edge[i], cost[i]);
	}
}

inline void pushup(int x)
{
	mx[x] = mmax(mx[ls], mx[rs]);
	mn[x] = mmin(mn[ls], mn[rs]);
}

inline void realpd(int x, int val)
{
	mn[x] = mmin(mn[x], val);
	if (tag[x] == -1) tag[x] = val;
	else tag[x] = mmin(tag[x], val);
}

inline void pushdown(int x)
{
	if (x > (maxn << 2) || tag[x] == -1) return;
	realpd (ls, tag[x]);
	realpd (rs, tag[x]);
	tag[x] = -1;
}

void build (int l, int r, int x = 1)
{
	tag[x] = -1;
	if (l == r) {
		mx[x] = len[l];
		mn[x] = inf;
		return ;
	}
	int mid = (l + r) >> 1;
	build (l, mid, ls);
	build (mid + 1, r, rs);
	pushup(x);
}

int query(int l, int r, int tl, int tr, int val, int x = 1)
{
	if (l >= tl && r <= tr) {
		realpd(x, val);
		return mx[x];
	}
	pushdown(x);
	int mid = (l + r) >> 1, ans(-1);
	if (tl <= mid) ans = mmax(ans, query (l, mid, tl, tr, val, ls));
	if (tr > mid) ans = mmax(ans, query (mid + 1, r, tl, tr, val, rs));
	pushup(x);
	return ans;
}

int querymn(int l, int r, int tl, int tr, int x = 1)
{
	if (l >= tl && r <= tr) return mn[x];
	pushdown(x);
	int mid = (l + r) >> 1, ans(inf);
	if (tl <= mid) ans = mmin(ans, querymn (l, mid, tl, tr, ls));
	if (tr > mid) ans = mmin(ans, querymn (mid + 1, r, tl, tr, rs));
	return ans;
}


inline int query(int u, int v, int val)
{
	int ans(-1);
	while (top[u] != top[v]) {
		if (dep[top[u]] < dep[top[v]]) mswap(u, v);
		ans = mmax(ans, query (1, n, id[top[u]], id[u], val));
		u = f[top[u]];
	}
	if (dep[u] > dep[v]) mswap (u, v);
	if (u != v) ans = mmax(ans, query(1, n, id[u] + 1, id[v], val));
	return ans;
}

inline int querymn(int u, int v)
{
	int ans(inf);
	while (top[u] != top[v]) {
		if (dep[top[u]] < dep[top[v]]) mswap(u, v);
		ans = mmin(ans, querymn (1, n, id[top[u]], id[u]));
		u = f[top[u]];
	}
	if (dep[u] > dep[v]) mswap (u, v);
	if (u != v) ans = mmin(ans, querymn (1, n, id[u] + 1, id[v]));
	return ans;
}

inline void addedge(int u, int v, int w)
{
	nxt[++cur] = head[u];
	head[u] = cur;
	edge[cur] = v;
	cost[cur] = w;
}

inline void MST()
{
	sort(Lines + 1, Lines + m + 1);
	for (int i = 1; i <= n; ++i) f[i] = i;
	for (int i = 1; i <= m; ++i) {
		int x = Lines[i].u, y = Lines[i].v;
		int fx = find(x), fy = find(y);
		if (fx == fy) continue;
		vis[i] = 1;
		f[fx] = fy;
		addedge (x, y, Lines[i].w);
		addedge (y, x, Lines[i].w);
	}

	dfs(1, 0), dfs(1, 1, 0);
	build (1, n);
}

int main()
{
	freopen ("easy.in", "r", stdin);
	freopen ("easy.out", "w", stdout);
	n = read(), m = read();
	for (int i = 1; i <= m; ++i) {
		Lines[i].u = read();
	   	Lines[i].v = read();
		Lines[i].w = read();
		Lines[i].p = i;
	}

	MST();

	for (int i = 1; i <= m; ++i) {
		if (vis[i]) continue;
		ans[Lines[i].p] = query(Lines[i].u, Lines[i].v, Lines[i].w);
	}

	for (int i = 1; i <= m; ++i) {
		if (!vis[i]) continue;
		ans[Lines[i].p] = querymn (Lines[i].u, Lines[i].v);
	}

	for (int i = 1; i <= m; ++i) {
		printf ("%d\n", (ans[i] == -1 || ans[i] == inf) ? 1000000000 : ans[i]);
	}
}

T3

还没有看题， 鸽了~