JZOJ 4611. 【NOI2016模拟7.11】接水问题 （贪心+A*+可持久化线段树）

Description:

https://gmoj.net/senior/#main/show/4611

题解：

先把A从大到小排序，最小的由排序不等式显然。

考虑类似第k短路的A*的做法。

定义状态为一个已经确定的前缀，它自己的代价显然，它的估价函数为把剩下的数字从小到大填的代价。

以自己代价+估价函数代价放入堆里一直扩展下一个即可，队列中会有(n^2k)个，加上求代价的复杂度，时间复杂度：(O(n^3k))。

考虑优化扩展，注意假设要在下一位放数字，肯定是从小到大放，所以优化这个无用的扩展，可以减小一个n。

再考虑，一个状态一直放最小的到长度为n为止，这中间经过了很多状态，考虑把这些也优化了，

对状态重新定义为已知道了(p[1..n])的值，前(t-1)个固定了，第(t)个可以变大，第(t)个现在是第(w)大，(bz[t+1..n-1])表示(p[x])能不能和(p[x+1])交换。

发现状态只有(O(k))个了，时间复杂度：(O(nk))。

用可持久化线段树维护即可做到(O(n~log~n+k~log~n))

Code：

#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("
")
using namespace std;

const ll inf = 1e18;

const int N = 2e5 + 5;

int n, k; ll a[N];

int cmpa(int x, int y) {
	return x > y;
}

ll sum;

#define i0 t[i].l
#define i1 t[i].r

#define pii pair<ll, int>
#define fs first
#define se second

struct tree {
	int l, r, p;
	bool ban, lz;
	pii x, y;
} t[N * 100]; int tt;

void upd(int i) {
	t[i].x = min(t[i0].x, t[i1].x);
	t[i].y = min(t[i0].y, t[i1].y);
}

void bt(int &i, int x, int y) {
	i = ++ tt;
	if(x == y) {
		t[i].p = x;
		t[i].x = t[i].y = pii(a[x] - a[x + 1], x);
		return;
	}
	int m = x + y >> 1;
	bt(i0, x, m); bt(i1, m + 1, y);
	upd(i);
}

int pl, pr, px, py, pz;

void dgp(int &i, int x, int y) {
	if(x == y) { px = t[i].p; return;}
	int m = x + y >> 1;
	if(pl <= m) dgp(i0, x, m); else dgp(i1, m + 1, y);
}
int fp(int &i, int x) {
	pl = pr = x;
	dgp(i, 1, n);
	return px;
}

void kq(int &i) {
	if(i) {
		t[++ tt] = t[i]; i = tt;
		t[i].ban = 0;
		t[i].lz = 1;
		t[i].x = t[i].y;
	}
}
void down(int &i) {
	if(t[i].lz) {
		kq(i0), kq(i1);
		t[i].lz = 0;
	}
}

void dd(int &i, int x, int y) {
	if(y < pl || x > pr) return;
	t[++ tt] = t[i], i = tt;
	if(x == y) {
		t[i].y.fs += (ll) px * (a[x] - a[x + 1]);
		if(t[i].ban) t[i].x = pii(inf, x); else
			t[i].x = t[i].y;
		if(py) t[i].p = pz;
		return;
	}
	int m = x + y >> 1; down(i);
	dd(i0, x, m); dd(i1, m + 1, y);
	upd(i);
}
void xiu(int &i, int x, int y) {
	int v = fp(i, x);
	if(x > 1) {
		py = 0;
		pl = pr = x - 1, px = y - v;
		dd(i, 1, n);
	}
	py = 1; pz = y;
	pl = pr = x, px = v - y;
//	pp("xiu %d %d %d %d
", x, y, v, px);
	dd(i, 1, n);
}

void du(int &i, int x, int y) {
	if(y < pl || x > pr) return;
	t[++ tt] = t[i], i = tt;
	if(x == y) {
		t[i].ban = 1;
		t[i].x = pii(inf, x);
		return;
	}
	int m = x + y >> 1; down(i);
	du(i0, x, m); du(i1, m + 1, y);
	upd(i);
}
void jz(int &i, int x) {
	pl = pr = x;
	du(i, 1, n);
}

void ddq(int &i, int x, int y) {
	if(y < pl || x > pr) return;
	if(x == y) {
		px = t[i].ban; return;
	}
	int m = x + y >> 1; down(i);
	ddq(i0, x, m); ddq(i1, m + 1, y);
}
int qry_ban(int &i, int x) {
	pl = pr = x;
	ddq(i, 1, n);
	return px;
}

pii pu;

void ft(int &i, int x, int y) {
	if(y < pl || x > pr) return;
	if(x >= pl && y <= pr) {
		pu = min(pu, t[i].x);
		return;
	}
	int m = x + y >> 1; down(i);
	ft(i0, x, m); ft(i1, m + 1, y);
}

pii qry(int &i, int t) {
	pl = t + 1, pr = n - 1; 
	pu = pii(inf, 0);
	ft(i, 1, n);
	return pu;
}

int rt;

struct nod {
	int g, t, w;
	pii c;
	ll s;
};

bool operator < (nod a, nod b) {
	return a.s > b.s;
}

priority_queue<nod> q;

void build() {
	bt(rt, 1, n);
	
	nod b;
	b.g = rt; b.t = 1; b.w = 2;
	b.c = pii(a[1] - a[2], 1);
	b.c = min(b.c, qry(b.g, b.t));
	b.s = sum;
	b.s += b.c.fs;
	
	q.push(b);
}

void swap(nod &b, int x, int y) {
	int v1 = fp(b.g, x), v2 = fp(b.g, y);
//	pp("  swap %d %d %d %d
", x, y, v1, v2);
//	pp("%lld
", qry(b.g, 1).fs);
	xiu(b.g, x, v2); xiu(b.g, y, v1);
//	pp("%lld
", qry(b.g, 1).fs);
}

int mak(nod &b) {
	if(b.w > n) {
		b.t ++; b.w = b.t + 1;
		if(b.t >= n) return 0;
	}
	b.c = pii(inf, 0);
	if(!qry_ban(b.g, b.t)) {
		b.c = pii((a[b.t] - a[b.w]) * (fp(b.g, b.w) - fp(b.g, b.t)), b.t);
	}
	b.c = min(b.c, qry(b.g, b.t));
	if(b.c.fs == inf) return 0;
	
	b.s += b.c.fs;
	return 1;
}

void work(nod b) {
	int x = b.c.se, y = x == b.t ? b.w : x + 1;
	if(x == b.t) {
		nod c = b;	
		swap(c, x, y);
		c.w ++; kq(c.g);
		if(mak(c)) q.push(c);
		
		
		b.s -= b.c.fs;
		jz(b.g, x);
		if(mak(b)) q.push(b);
	} else {
		nod c = b;
		swap(c, x, y);
		c.t = x, c.w = x + 2;
		kq(c.g);
		if(mak(c)) q.push(c);
		
		b.s -= b.c.fs;
		jz(b.g, x);
		if(mak(b)) q.push(b);
	}
}

int main() {
	freopen("water.in", "r", stdin);
	freopen("water.out", "w", stdout);
	scanf("%d %d", &n, &k);
	fo(i, 1, n) scanf("%lld", &a[i]);
	sort(a + 1, a + n + 1, cmpa);
	fo(i, 1, n)	sum += a[i] * i;
	build();
	pp("%lld
", sum);
	fo(ii, 1, k - 1) {
		nod b = q.top(); q.pop();
		pp("%lld
", b.s);
		work(b);
//		pp("fs = %lld
", qry(b.g, b.t).fs);
//		pp("%d %d %lld %d
", b.t, b.w, b.c.fs, b.c.se);
//		fo(i, 1, n) {
//			int x = b.c.se, y = x == b.t ? b.w : x + 1;
//			int t = i == x ? y : (i == y ? x : i);
//			pp("%d ", fp(b.g, t));
//		}
//		hh;
	}
}