【BZOJ3218】【UOJ#77】a + b Problem

题目

题目在这里

思路&做法

明显的最小割（其实是之前做过一道类似的题）

1. S向每一个格子连容量为\(b_i\)的边

2. 每一个格子向T连容量为\(w_i\)的边

3. 对于格子\(i\)向满足条件的格子\(j(1 \leq j < i, l_i \leq a_j \leq r_i)\)连容量为\(p_i\)的边

但是考虑到这题恶心的数据范围， 这样做很明显会TLE。

算法的瓶颈在第三步。我们发现\(j\)的范围看起来像是一个像二维偏序的东西， 于是便可以用主席树来优化一下， 对于每个节点\(i\)新建一个节点\(k\)，由i向k连容量为\(p_i\)的边， 由\(k\)向主席树中\(j\)的范围的对应的节点连容量为\(inf\)的边。

代码

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
const int INF = 0x7F7F7F7F;
const int N = 50010, M = 1450010;
struct edge
{	int from, to, flow, cap;
	edge() { }
	edge(int _1, int _2, int _3, int _4) : from(_1), to(_2), flow(_3), cap(_4) { }
};
struct Dinic
{	edge edges[M];
	int head[165022], nxt[M], tot;
	int L, R;
	inline void init()
	{	memset(head, -1, sizeof(head));
		tot = 0;
	}
	void add_edge(int x, int y, int z)
	{	edges[tot] = edge(x, y, 0, z);
		nxt[tot] = head[x];
		head[x] = tot++;
		edges[tot] = edge(y, x, 0, 0);
		nxt[tot] = head[y];
		head[y] = tot++;
	}
	int s, t;
	int d[165022];
	bool bfs()
	{	memset(d, -1, sizeof(d));
		queue<int> q;
		d[s] = 0;
		q.push(s);
		while (!q.empty())
		{	int x = q.front(); q.pop();
			for (int i = head[x]; ~i; i = nxt[i])
			{	edge & e = edges[i];
				if (e.cap > e.flow && d[e.to] == -1)
				{	d[e.to] = d[x] + 1;
					q.push(e.to);
				}
			}
		}
		return d[t] != -1;
	}
	int cur[165022];
	int dfs(int x, int a)
	{	if (x == t || a == 0) return a;
		int flow = 0, f;
		for (int & i = cur[x]; ~i; i = nxt[i])
		{	edge & e = edges[i];
			if (d[e.to] == d[x] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0)
			{	e.flow += f;
				edges[i^1].flow -= f;
				a -= f;
				flow += f;
				if (a == 0) break;
			}
		}
		return flow;
	}
	int maxflow(int _s, int _t)
	{	s = _s, t = _t;
		int flow = 0;
		while (bfs())
		{	for (int i = L; i <= R; i++)
				cur[i] = head[i];
			flow += dfs(s, INF);
		}
		return flow;
	}
} dinic;
int n;
int root[N];
struct SegmentTree
{	int ls[N*20], rs[N*20], sz;
	void update(int & cur, int pre, int l, int r, int p, int node)
	{	cur = ++sz;
		if (l == r)
		{	if (pre) dinic.add_edge(cur, pre, INF);
			dinic.add_edge(cur, node, INF);
			return;
		}
		ls[cur] = ls[pre];
		rs[cur] = rs[pre];
		int mid = (l + r) >> 1;
		if (p <= mid)
			update(ls[cur], ls[pre], l, mid, p, node);
		else
			update(rs[cur], rs[pre], mid+1, r, p, node);
		dinic.add_edge(cur, ls[cur], INF);
		dinic.add_edge(cur, rs[cur], INF);
	}
	void link(int cur, int l, int r, int ql, int qr, int node)
	{	if (ql == l && qr == r)
			dinic.add_edge(node, cur, INF);
		else
		{	int mid = (l + r) >> 1;
			if (qr <= mid)
				link(ls[cur], l, mid, ql, qr, node);
			else if (ql > mid)
				link(rs[cur], mid+1, r, ql, qr, node);
			else
				link(ls[cur], l, mid, ql, mid, node),
				link(rs[cur], mid+1, r, mid+1, qr, node);
		}
	}
} st;
int a[N], w[N], b[N], L[N], R[N], p[N];
int num[N], total;
int main()
{	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{	scanf("%d %d %d %d %d %d", &a[i], &b[i], &w[i], &L[i], &R[i], &p[i]);
		num[++total] = a[i];
		num[++total] = L[i];
		num[++total] = R[i];
	}
	sort(num+1, num+total+1);
	total = unique(num+1, num+total+1) - num - 1;
	for (int i = 1; i <= n; i++)
	{	a[i] = lower_bound(num+1, num+1+total, a[i]) - num;
		L[i] = lower_bound(num+1, num+1+total, L[i]) - num;
		R[i] = lower_bound(num+1, num+1+total, R[i]) - num;
	}
	int ans = 0;
	dinic.init();
	int S = n + n + 1, T = n + n + 2;
	st.sz = n + n + 2;
	for (int i = 1; i <= n; i++)
	{	dinic.add_edge(S, i, b[i]);
		dinic.add_edge(i, T, w[i]);
		dinic.add_edge(i, i+n, p[i]);
		if (i > 1)
			st.link(root[i-1], 1, total, L[i], R[i], i+n);
		st.update(root[i], root[i-1], 1, total, a[i], i);
		ans += b[i] + w[i];
	}
	dinic.L = 0, dinic.R = st.sz;
	ans -= dinic.maxflow(S, T);
	printf("%d\n", ans);
}
/*
10
0 1 7 3 9 2
7 4 0 9 10 5
1 0 4 2 10 2
7 9 1 5 7 2
6 3 5 3 6 2
6 6 4 1 8 1
6 1 6 0 6 5
2 2 5 0 9 3
5 1 3 0 2 5
5 6 7 1 1 2
*/