Network POJ - 3417 LCA+树上差分

Yixght is a manager of the company called SzqNetwork(SN). Now she’s very worried because she has just received a bad news which denotes that DxtNetwork(DN), the SN’s business rival, intents to attack the network of SN. More unfortunately, the original network of SN is so weak that we can just treat it as a tree. Formally, there are N nodes in SN’s network, N-1 bidirectional channels to connect the nodes, and there always exists a route from any node to another. In order to protect the network from the attack, Yixght builds M new bidirectional channels between some of the nodes.

As the DN’s best hacker, you can exactly destory two channels, one in the original network and the other among the M new channels. Now your higher-up wants to know how many ways you can divide the network of SN into at least two parts.
Input
The first line of the input file contains two integers: N (1 ≤ N ≤ 100 000), M (1 ≤ M ≤ 100 000) — the number of the nodes and the number of the new channels.

Following N-1 lines represent the channels in the original network of SN, each pair (a,b) denote that there is a channel between node a and node b.

Following M lines represent the new channels in the network, each pair (a,b) denote that a new channel between node a and node b is added to the network of SN.
Output

Output a single integer — the number of ways to divide the network into at least two parts.

Sample Input

4 1
1 2
2 3
1 4
3 4

Sample Output

3

---

lca + 差分即可；
其中 fg 表示该边被覆盖多少次；
那么显然我们只能选 0 或 1 次的边；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10


inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

int head[maxn], nxt[maxn], ver[maxn];
int cnt;
int n, m, N;


void addedge(int x, int y) {
	ver[++cnt] = y; nxt[cnt] = head[x]; head[x] = cnt;
}

int grand[maxn][20];
int depth[maxn];
int fg[maxn];

void init() {
	memset(head, 0, sizeof(head));
	memset(fg, 0, sizeof(fg));
	N = (int)(log(n) / log(2)) + 1;
	depth[1] = 1; cnt = 0;
}
void dfs(int rt) {
	for (int i = 1; i <= N; i++) {
		grand[rt][i] = grand[grand[rt][i - 1]][i - 1];
	}
	for (int i = head[rt]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == grand[rt][0])continue;
		grand[v][0] = rt;
		depth[v] = depth[rt] + 1;
		dfs(v);
	}
}

int lca(int a, int b) {
	if (depth[a] > depth[b])swap(a, b);
	for (int i = N; i >= 0; i--) {
		if (depth[a] <= depth[grand[b][i]])
			b = grand[b][i];
	}
	if (a == b)return a;
	for (int i = N; i >= 0; i--) {
		if (grand[a][i] != grand[b][i]) {
			a = grand[a][i]; b = grand[b][i];
		}
	}
	return grand[a][0];
}

void DFS(int rt) {
	for (int i = head[rt]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == grand[rt][0])continue;
		DFS(v);
		fg[rt] += fg[v];
	}
}



int main()
{
	//ios::sync_with_stdio(false);
	while (cin >> n >> m) {
		init();
		for (int i = 1; i < n; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			addedge(x, y); addedge(y, x);
		}
		dfs(1);
		for (int i = 0; i < m; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			fg[x]++; fg[y]++;
			fg[lca(x, y)] -= 2;
		}
		DFS(1);
		long long ans = 0;
		for (int i = 2; i <= n; i++) {
			if (fg[i] == 0)ans += m;
			else if (fg[i] == 1)ans++;
		}
		cout << ans << endl;
	}
}