[AHOI2005]航线规划

【题目链接】
洛谷 P2542 [AHOI2005]航线规划
BZOJ 1969: [Ahoi2005]LANE 航线规划

【分析】

• 删边很麻烦，考虑离线后的倒序加边操作
• 然而还是很难做。。。
• 从特殊情况入手，考虑图是一棵树的情况
• 发现加边与将树上关键路径更改为非关键路径是等价的
• 将关键路径看做边权为$0$的边，非关键路径看做边权为$1$的边
• 现在问题转化为在树上对一段路径赋值和查询一段路径的边权和
• 在原图上随机生成一棵树，用树链剖分维护

【代码】

#include
#define seg 1, 1, n
using namespace std;

const int maxn = 3e4 + 10;
const int maxm = 1e5 + 10;

int n, m;
int ver[maxn], sumv[maxn << 2], setv[maxn << 2], f[maxn], ans[maxn << 1];
int dfn[maxn], fa[maxn], top[maxn], size[maxn], son[maxn], dep[maxn];
bool intree[maxm], erase[maxm];
struct edge0
{
	int u, v;
	inline bool operator < (const edge0 &a) const
	{
		if(u == a.u)
			return v < a.v;
		return u < a.u;
	}
} e0[maxm];
map<edge0, int> mp;
struct edge
{
	int v, nxt;
} e[maxn << 1];
int qu;
struct query
{
	int c;
	int u, v;
	int x;
} q[maxn << 1];

inline void adde(int u, int v)
{
	static int ed = 1;
	e[++ed] = (edge){ v, ver[u] };
	ver[u] = ed;
	//cout << ed << endl;
}

inline void dfs1(int u, int f)
{
	size[u] = 1, fa[u] = f;
	for(int i = ver[u]; i; i = e[i].nxt)
	{
		int v = e[i].v;
		if(v == f)
			continue;
		dep[v] = dep[u] + 1;
		dfs1(v, u);
		size[u] += size[v];
		if(size[son[u]] < size[v])
			son[u] = v;
	}
}

inline void dfs2(int u, int t)
{
	static int cnt = 0;
	dfn[u] = ++cnt, top[u] = t;
	if(!son[u])
		return;
	dfs2(son[u], t);
	for(int i = ver[u]; i; i = e[i].nxt)
	{
		int v = e[i].v;
		if(dfn[v])
			continue;
		dfs2(v, v);
	}
}

inline void up(int o, int lc, int rc)
{
	sumv[o] = sumv[lc] + sumv[rc];
}

inline void down(int o, int lc, int rc)
{
	sumv[lc] = sumv[rc] = 0;
	setv[lc] = setv[rc] = 1;
	setv[o] = 0;
}

inline void build(int o, int l, int r)
{
	if(l == r)
	{
		sumv[o] = 1;
		return;
	}
	int lc = o << 1, rc = lc | 1, mid = (l + r) >> 1;
	build(lc, l, mid);
	build(rc, mid + 1, r);
	up(o, lc, rc);
}

inline void set0(int s, int t, int o, int l, int r)
{
	if(l == r)
	{
		sumv[o] = 0;
		setv[o] = 1;
		return;
	}
	int lc = o << 1, rc = lc | 1, mid = (l + r) >> 1;
	if(setv[o])
		down(o, lc, rc);
	if(s <= mid)
		set0(s, t, lc, l, mid);
	if(mid < t)
		set0(s, t, rc, mid + 1, r);
	up(o, lc, rc);
}

inline int sums(int s, int t, int o, int l, int r)
{
	if(s <= l && r <= t)
		return sumv[o];
	int lc = o << 1, rc = lc | 1, mid = (l + r) >> 1, ans = 0;
	if(setv[o])
		down(o, lc, rc);
	if(s <= mid)
		ans += sums(s, t, lc, l, mid);
	if(mid < t)
		ans += sums(s, t, rc, mid + 1, r);
	return ans;
}

inline void clear(int u, int v)
{
	while(top[u] != top[v])
	{
		if(dep[top[u]] < dep[top[v]])
			swap(u, v);
		set0(dfn[top[u]], dfn[u], seg);
		u = fa[top[u]];
	}
	if(u == v)
		return;
	if(dep[u] > dep[v])
		swap(u, v);
	set0(dfn[u] + 1, dfn[v], seg);
}

inline int ask(int u, int v)
{
	int ans = 0;
	while(top[u] != top[v])
	{
		if(dep[top[u]] < dep[top[v]])
			swap(u, v);
		ans += sums(dfn[top[u]], dfn[u], seg);
		u = fa[top[u]];
	}
	if(u == v)
		return ans;
	if(dep[u] > dep[v])
		swap(u, v);
	ans += sums(dfn[u] + 1, dfn[v], seg);
	return ans;
}

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

int main()
{
	scanf("%d%d", &n, &m);
	
	for(int i = 1; i <= n; i++)
		f[i] = i;
	
	for(int i = 1; i <= m; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		if(u > v)
			swap(u, v);
		e0[i].u = u, e0[i].v = v;
		mp[(edge0){ u, v }] = i;
		int fu = find(u), fv = find(v);
		if(fu == fv)
			continue;
		f[fv] = fu;
		intree[i] = true;
		adde(u, v);
		adde(v, u);
	}
	
	dfs1(1, 0);
	dfs2(1, 1);
	build(seg);
	
	int c;
	scanf("%d", &c);
	while(c != -1)
	{
		q[++qu].c = c;
		int u, v;
		scanf("%d%d", &u, &v);
		if(u > v)
			swap(u, v);
		q[qu].u = u, q[qu].v = v;
		int x = mp[(edge0){ u, v }];
		q[qu].x = x;
		if(c == 0)
			erase[x] = true;
		scanf("%d", &c);
	}
	
	for(int i = 1; i <= m; i++)
	{
		if(intree[i] || erase[i])
			continue;
		clear(e0[i].u, e0[i].v);
	}
	
	memset(ans, -0x3f, sizeof ans);
	for(int i = qu; i > 0; i--)
	{
		if(q[i].c == 1)
			ans[i] = ask(q[i].u, q[i].v);
		else
			clear(q[i].u, q[i].v);
	}
	
	for(int i = 1; i <= qu; i++)
	{
		if(ans[i] < 0)
			continue;
		printf("%d\n", ans[i]);
	}
	
	/*
	for(int i = 1; i <= qu; i++)
		cout << q[i].x << " ";
	*/
	
	return 0;
}