CERC 2014 (动态树+主席树）

CERC 2014 Pork barrel

Problem :
n个点m条边有边权的无向图，有q个询问，每次询问权值在[L，R]内的边组成的最小生成树的权值和，强制在线。
n <= 1000, m <= 100000, q <= 100000
Solution :
参考了网上的一份题解
按照边权从大到小加入边，用LCT来维护最小生成树。再用一棵权值主席树，第i棵主席树记录表示权值大于等于 i 的边所构成的最小生成树边权和。
对于每个询问[L, R]直接在第L棵主席树的[L ,R]区间内统计答案。

对于每个询问[L, R]，要将端点离散化成对应的边权表示，要注意离散化后的区间应被原来的区间包含，而不是包含原来的区间。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <queue>
#include <cassert>
using namespace std;
#define f(i, x, y) for (int i = x; i <= y; ++i)
#define fd(i, x, y) for (int i = x; i >= y; --i)
#define rep(i, x, y) for (int i = x; i <= y; ++i)
#define repd(i, x, y) for (int i = x; i >= y; --i)

const int INF = 1e9 + 7;
const int N = 300008;
const int NN = N * 100;
int n, m, q;

void read(int &x)
{
	char ch;
	for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar());
	x = 0;
	for (; ch >= '0' && ch <= '9'; ch = getchar())
		 x = x * 10 + ch -  '0';
}
struct edge
{
	int u, v, w;
	bool operator < (const edge &b) const
	{
		return w > b.w;
	}
}eg[N];
int fa[N], c[N][2], val[N], mx[N], rev[N], st[N];
int root[N], rtId[N], ls[NN], rs[NN];
long long tag[NN];
int num, total;
int p[N];
int tot;

bool isroot(int x)
{
	return c[fa[x]][0] != x && c[fa[x]][1] != x;
}
void pushup(int x)
{
	int l = c[x][0], r = c[x][1];
	mx[x] = x;
	if (val[mx[l]] > val[mx[x]]) mx[x] = mx[l];
	if (val[mx[r]] > val[mx[x]]) mx[x] = mx[r];
}
void pushdown(int x)
{
	int l = c[x][0], r = c[x][1];
	if (rev[x])
	{
		if (l) rev[l] ^= 1;
		if (r) rev[r] ^= 1;
		rev[x] ^= 1;
		swap(c[x][0], c[x][1]);
	}
}
void rotate(int x)
{
	int y = fa[x], z = fa[y], l, r;
	if (c[y][0] == x) l = 0; else l = 1; r = l ^ 1;
	if (!isroot(y))
	{
		if (c[z][0] == y) c[z][0] = x; else c[z][1] = x;
	}
	fa[x] = z; fa[y] = x; fa[c[x][r]] = y;
	c[y][l] = c[x][r]; c[x][r] = y;
	pushup(y); pushup(x);
}
void splay(int x)
{
	int top = 0; st[++top] = x;
	for (int i = x; !isroot(i); i = fa[i]) st[++top] = fa[i];
	while (top) pushdown(st[top--]);
	while (!isroot(x))
	{
		int y = fa[x], z = fa[y];
		if (!isroot(y))
		{
			if (c[y][0] == x ^ c[z][0] == y) rotate(x);
			else rotate(y);
		}
		rotate(x);
	}
}
void access(int x)
{
	for (int t = 0; x; t = x, x = fa[x])
	{
		splay(x); 
		c[x][1] = t;
		pushup(x);
	}
}
void rever(int x)
{
	access(x); splay(x); rev[x] ^= 1;
}
void link(int u, int v)
{
	rever(u); fa[u] = v;
}
void cut(int u, int v)
{
	rever(u); access(v); splay(v); fa[c[v][0]] = 0; c[v][0] = 0; pushup(v);
}
int find(int u)
{
	access(u); splay(u);
	while (c[u][0]) u = c[u][0];
	return u;
}
int query(int u, int v)
{
	rever(u); access(v); splay(v); return mx[v];
}
void build(int &rt, int l, int r)
{
	rt = ++total;
	ls[rt] = rs[rt] = tag[rt] = 0;
	if (l == r) return;
	int m = l + r >> 1;
	build(ls[rt], l, m);
	build(rs[rt], m + 1, r);
}
void insert(int &rt, int last, int pos, int val, int l, int r)
{
	rt = ++total;
	ls[rt] = ls[last]; rs[rt] = rs[last]; tag[rt] = tag[last];
	if (l == r)
	{
		tag[rt] += val;
		return;
	}
	int m = l + r >> 1;
	if (pos <= m) insert(ls[rt], ls[last], pos, val, l, m);
	if (m <  pos) insert(rs[rt], rs[last], pos, val, m + 1, r);
	tag[rt] = tag[ls[rt]] + tag[rs[rt]];
}
long long query(int rt, int L, int R, int l, int r)
{
	if (L <= l && r <= R)
	{
		return tag[rt];
	}
	long long res = 0;
	int m = l + r >> 1;
	if (L <= m) res += query(ls[rt], L, R, l, m);
	if (m <  R) res += query(rs[rt], L, R, m + 1, r);
	return res;
}
void init()
{
	read(n); read(m);
	for (int i = 1; i <= m; ++i) 
	{
		read(eg[i].u); read(eg[i].v); read(eg[i].w);
		p[i] = eg[i].w;
	}
	sort(p + 1, p + m + 1);
	tot = unique(p + 1, p + m + 1) - p - 1;
	for (int i = 1; i <= m; ++i)
		eg[i].w = lower_bound(p + 1, p + tot + 1, eg[i].w) - p;
}
void clear()
{
	for (int i = 1; i <= num; ++i) root[i] = 0;
	for (int i = 1; i <= tot + 5; ++i) rtId[i] = 0;
	for (int i = 1; i <= n + m; ++i)
	{
		fa[i] = c[i][0] = c[i][1] = val[i] = mx[i] = rev[i] = 0;
	}
	num = total = 0;
}
void work()
{
	build(root[0], 1, tot);
	sort(eg + 1, eg + m + 1);
	for (int i = 1; i <= m; ++i)
	{
		int u = eg[i].u, v = eg[i].v, w = eg[i].w;
		if (find(u) == find(v))
		{
			int t = query(u, v);
			cut(t, eg[t - n].u);
			cut(t, eg[t - n].v);
			rtId[w] = ++num;
			insert(root[num], root[num - 1], val[t], -p[val[t]], 1, tot);	
		}
		val[i + n] = w; mx[i + n] = i + n;
		link(i + n, u);
		link(i + n, v);
		rtId[w] = ++num;
		insert(root[num], root[num - 1], w, p[w], 1, tot);
	}
}
void solve()
{		
	read(q);
	int ans = 0;
	for (int i = 1; i <= q; ++i)
	{
		int u, v; 
		read(u); read(v);
		u -= ans; v -= ans;
		int l = lower_bound(p + 1, p + tot + 1, u) - p;
		int r = upper_bound(p + 1, p + tot + 1, v) - p - 1;	
		if (r == tot + 1) r = tot;	
		ans = query(root[rtId[l]], l, r, 1, tot);
		cout << ans << endl;
	}
}
int main()
{	
	int T; read(T);
	for (int cas = 1; cas <= T; ++cas)
	{
		init();
		clear();
		work();
		solve();
	}
}