【BZOJ3772】精神污染

题目

传送门

题解&算法

看完题后千万不要被概率吓到

这题的难点在于求与一条路径被另一条路径包含的总数量

很明显如果以x,a被x,b包含， 那么\(in[b] <= in[a] < out[b]\)， 且\(out[a] > out[b]\)

于是我们可以用主席树来处理一下， 建树时差分一下， 查询时也是前缀和差分

代码

#include <iostream>
#include <cstdlib>
#include <vector>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 100010, M = 200010;

typedef long long LL;

struct edge
{	int from, to;
	edge() { }
	edge(int _1, int _2) : from(_1), to(_2) { }
} edges[M];

int head[N], nxt[M], tot;

inline void init()
{	memset(head, -1, sizeof(head));
	tot = 0;
}

inline void add_edge(int x, int y)
{	edges[tot] = edge(x, y);
	nxt[tot] = head[x];
	head[x] = tot++;
	edges[tot] = edge(y, x);
	nxt[tot] = head[y];
	head[y] = tot++;
}

int dep[N], father[N][17];
int in[N], out[N], dfs_clock;

void dfs(int x)
{	for (int i = 1; (1<<i) <= dep[x]; i++)
		father[x][i] = father[father[x][i-1]][i-1];
	in[x] = ++dfs_clock;
	for (int i = head[x]; ~i; i = nxt[i])
	{	edge & e = edges[i];
		if (e.to != father[x][0])
		{	dep[e.to] = dep[x] + 1;
			father[e.to][0] = x;
			dfs(e.to);
		}
	}
	out[x] = ++dfs_clock; //少了++样例输出就变成-1/1了
}

int getLCA(int x, int y)
{	if (dep[x] > dep[y]) swap(x, y);
	int d = dep[y] - dep[x];
	for (int i = 0; (1<<i) <= d; i++)
		if (d & (1<<i)) y = father[y][i];
	if (x == y) return x;
	for (int i = 16; i >= 0; i--)
	{	if (father[x][i] != father[y][i])
		{	x = father[x][i];
			y = father[y][i];
		}
	}
	return father[x][0]; 
}

int root[N * 2];

struct SegmentTree
{	int ls[3800010], rs[3800010], sz;
	int sumv[3800010];//开40000010会MLE

	inline void cpynode(int x, int y)
	{	ls[x] = ls[y];
		rs[x] = rs[y];
		sumv[x] = sumv[y];
	}

	void insert(int & cur, int pre, int l, int r, int p, int val)
	{	cur = ++sz;
		cpynode(cur, pre);
		sumv[cur] +=  val;
		if (l == r) return;
		int mid = (l + r) >> 1;
		if (p <= mid)
			insert(ls[cur], ls[pre], l, mid, p, val);
		else
			insert(rs[cur], rs[pre], mid+1, r, p, val);
	}

	LL query(int x, int y, int z, int k, int l, int r, int ql, int qr)
	{	if (ql == l && qr == r)
			return sumv[x] + sumv[y] - sumv[z] - sumv[k];
		int mid = (l + r) >> 1;
		if (qr <= mid)
			return query(ls[x], ls[y], ls[z], ls[k], l, mid, ql, qr);
		else if (ql > mid)
			return query(rs[x], rs[y], rs[z], rs[k], mid+1, r, ql, qr);
		else
			return query(ls[x], ls[y], ls[z], ls[k], l, mid, ql, mid) + 
				   query(rs[x], rs[y], rs[z], rs[k], mid+1, r, mid+1, qr);
	}
} st;

LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a % b); }

int n, m;

struct Data
{	int x, y;
	Data() { }
	Data(int _1, int _2) : x(_1), y(_2) { }
} L[N];

bool cmp(Data a, Data b) { return a.x != b.x ? a.x < b.x : a.y < b.y; }

vector<int> vec[N];

void dfs_2(int x)
{	root[x] = root[father[x][0]];
	for (int i = 0; i < vec[x].size(); i++)
		st.insert(root[x], root[x], 1, dfs_clock, in[vec[x][i]], 1),
		st.insert(root[x], root[x], 1, dfs_clock, out[vec[x][i]], -1);
	for (int i = head[x]; ~i; i = nxt[i])
	{	edge & e = edges[i];
		if (e.to != father[x][0])
			dfs_2(e.to);
	}
}

int main()
{	scanf("%d %d", &n, &m);
	init();
	for (int i = 1; i < n; i++)
	{	int x, y;
		scanf("%d %d", &x, &y);
		add_edge(x, y);
	}
	father[1][0] = 0;
	dep[1] = 1;
	dfs(1);
	for (int i = 1; i <= m; i++)
		scanf("%d %d", &L[i].x, &L[i].y), 
		vec[L[i].x].push_back(L[i].y);
	dfs_2(1);
	sort(L + 1, L + m + 1, cmp);
	LL ans1 = 0, ans2 = (LL) m * (m-1) / 2;
	for (int i = 1; i <= m; i++)
	{	int lca = getLCA(L[i].x, L[i].y);
		ans1 += st.query(root[L[i].x], root[L[i].y], root[lca], root[father[lca][0]], 1, dfs_clock, in[lca], in[L[i].x]) + 
				st.query(root[L[i].x], root[L[i].y], root[lca], root[father[lca][0]], 1, dfs_clock, in[lca], in[L[i].y]) -  //刚开始减号打成加号调了很久啊。。。
				st.query(root[L[i].x], root[L[i].y], root[lca], root[father[lca][0]], 1, dfs_clock, in[lca], in[lca])
				 - 1;
	}
	int j;
	for (int i = 1; i <= m; i = j)
		for (j = i; L[i].x == L[j].x && L[i].y == L[j].y; j++)
			ans1 -= (LL) (j - i) * (j - i - 1) / 2//一定要减掉重复！！！！！！！！！！！！！
	LL d = gcd(ans1, ans2);
	ans1 /= d, ans2 /= d;
	printf("%lld/%lld", ans1, ans2);
	return 0;
}

恶心的地方

1. 卡空间
2. 路径有重复