POJ3417 Network

题目链接

翻译：一棵无根树，给出m条边，把这m条边连上，每次你能毁掉两条边，规定一条是树边，一条新边。问有多少种方案能使树断裂。


这题想一想就明白了：考虑每一条新边连接的两个点\(x\)和\(y\)，对于这两点间树上路径的边，如果没有和其他新边连成环，那么这条边删了后树就会断裂，否则就断不了。所以每一次我们把\(x\)到\(y\)在树上的路径标记一遍，最后统计每一条边的标记情况：如果没有标记，那么随时可以删；如果只标记了一次，那么必须和对应的新边一块删；如果被标记了多次，那么任何时刻都删不了。
记被标记了0次、1次、多次的边的数量为\(n1,n2,n3\)，那么答案就是\(n1 * m + n2\)。


至于边标记，那么就要用到树上差分，而且是边差分。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int N = 17;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
#endif
}

int n, m;
struct Edge
{
	int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y)
{
	e[++ecnt] = (Edge){head[x], y};
	head[x] = ecnt;
}

int fa[N + 2][maxn], dep[maxn];
In void dfs(int now, int _f)
{
	for(int i = 1; (1 << i) <= dep[now]; ++i)
		fa[i][now] = fa[i - 1][fa[i - 1][now]];
	forE(i, now, v)
	{
		if(v == _f) continue;
		dep[v] = dep[now] + 1, fa[0][v] = now;
		dfs(v, now);
	}
}
In int lca(int x, int y)
{
	if(dep[x] < dep[y]) swap(x, y);
	for(int i = N; i >= 0; --i)
		if(dep[fa[i][x]] >= dep[y]) x = fa[i][x];
	if(x == y) return x;
	for(int i = N; i >= 0; --i)
		if(fa[i][x] != fa[i][y]) x = fa[i][x], y = fa[i][y];
	return fa[0][x]; 
}

int a[maxn];
ll ans = 0;
In void dfs2(int now, int _f)
{
	forE(i, now, v)
	{
		if(v == _f) continue;
		dfs2(v, now);
		a[now] += a[v]; 
	}
	if(now == 1) return;
	if(!a[now]) ans += m;
	else if(a[now] == 1) ++ans;
}

int main()
{
//	MYFILE();
	Mem(head, -1), ecnt = -1; 
	n = read(), m = read();
	for(int i = 1; i < n; ++i)
	{
		int x = read(), y = read();
		addEdge(x, y), addEdge(y, x);
	}
	dep[1] = 1, dfs(1, 0);
	for(int i = 1; i <= m; ++i)
	{
		int x = read(), y = read();
		int z = lca(x, y);
		a[x]++, a[y]++, a[z] -= 2;
	}
	dfs2(1, 0);
	write(ans), enter;
	return 0;	
}