NOI2019 序列

link ### 题解

不考虑Ai,Bi的顺序，则相当于每次选出k个Pair(Ai,Bi)，使得至少l个Ai=Bi

考虑费用流建图。

(文中a,b的边表示容量为a，费用为-b)

建S0,T0两个点,S向S0连k,0的边，T0向T连k,0的边。

S0向Ai连1,ai的边，Bi向T0连1,bi的边

Ai向Bi连1,0的边（表示两边都指定）

对于只指定一边的情况，建S1,T1表示只指定Bi,Ai的情况，S1向T1连k-l,0的边，Ai向S1连1,0的边，T1向Bi连1,0的边。

考虑怎么优化求费用流的过程，就是贪心模拟这一过程。

如果S1,T1的边还能有流量，则找一条S->S0->Max(ai)->S1->T1->Max(bi)->T0->T的增广路。

否则，增广路可能形如S->S0->Ai->Bi->T0->T，也可能形如S->S0->Ai->Bi->T1->Max(bi)->T0->T(Bi->T0的边已经满了)，还可能形如S->S0->Max(ai)->S1->Ai->Bi->T0->T(S0->Ai的边已经满了)。

如果Ai->S1,T1->Bi都满了，那么就把Ai->S1->T1->Bi这条路径变成Ai->Bi。

用几个优先队列算一下每种情况就可以了

不过这样没法AC，只能84分(88?)，要AC需要把只有删除的优先队列变成一次排序+删除标记。

/*
Author: QAQ Automaton
Lang: C++
Prog: sequence.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#include<ext/pb_ds/priority_queue.hpp>
#define int ll
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	// input a signed integer
	inline bool read (signed &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}

	inline bool read (long long &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		x=gc();
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	// print a signed integer
	inline bool write (signed x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}

	inline bool write (long long x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int a[200005],b[200005];
int inf;
struct _init_{
	_init_(){
		memset(&inf,0x3f,sizeof(inf));
	}
};
_init_ __init__;
__gnu_pbds::priority_queue<pii> p2,p3;
__gnu_pbds::priority_queue<pii>::point_iterator i2[200005],i3[200005];
struct xqueue{
	pii a[200005];
	bool del[200005];
	int it;
	void erase(int x){
		del[x]=1;
	}
	void init(int n){
		for(int i=1;i<=n;++i)del[i]=0;
		sort(a+1,a+n+1,dcmp<pii>);
		it=1;
	}
	pii top(){
		while(del[a[it].y])++it;
		return a[it];
	}
}p1,p4,p5;
int sx[200005],sy[200005];
int t;
void selectX(int x){
	p4.erase(x);
	if(!sy[x]){
		p1.erase(x);
		i3[x]=p3.push(make_pair(b[x],x));
	}
	else{
		p2.erase(i2[x]);
		--t;
	}
	sx[x]=1;
}
void selectY(int x){
	p5.erase(x);
	if(!sx[x]){
		p1.erase(x);
		i2[x]=p2.push(make_pair(a[x],x));
	}
	else{
		p3.erase(i3[x]);
		--t;
	}
	sy[x]=1;
}
void selectXY(int x){
	sx[x]=sy[x]=1;
	p1.erase(x);
	p4.erase(x);
	p5.erase(x);
}
signed main(){
	freopen("sequence.in","r",stdin);
	freopen("sequence.out","w",stdout);
	int T,n,k,l;
	read(T);
	for(;T;--T){
		read(n,k,l);
		l=k-l;
		for(int i=1;i<=n;++i)read(a[i]);
		for(int i=1;i<=n;++i)read(b[i]);
		for(int i=1;i<=n;++i)sx[i]=sy[i]=0;
		p2.clear();
		p3.clear();
		for(int i=1;i<=n;++i){
			p1.a[i]=(make_pair(a[i]+b[i],i));
			p4.a[i]=(make_pair(a[i],i));
			p5.a[i]=(make_pair(b[i],i));
		}
		p1.init(n);
		p4.init(n);
		p5.init(n);
		int s=0;
		t=0;
		for(int i=1;i<=k;++i){
			if(t<l){
				++t;
				s+=p4.top().x;
				s+=p5.top().x;
				selectX(p4.top().y);
				selectY(p5.top().y);
			}
			else{
				int a=p1.top().x,b=p2.empty()?-inf:(p2.top().x+p5.top().x),c=p3.empty()?-inf:(p3.top().x+p4.top().x);
				if(a>b){
					if(a>c){
						selectXY(p1.top().y);
					}
					else{
						++t;
						selectY(p3.top().y);
						selectX(p4.top().y);
					}
				}
				else{
					if(b>c){
						++t;
						selectX(p2.top().y);
						selectY(p5.top().y);
					}else{
						++t;
						selectY(p3.top().y);
						selectX(p4.top().y);
					}
				}
				s+=max(a,max(b,c));
			}
		}
		write(s,'\n');
	}
	return 0;
}