CF618F Double Knapsack

Double Knapsack

CF618F (Luogu)

题面翻译

给你两个可重集 $A,B$，$A,B$ 的元素个数都为 $n$，它们中每个元素的大小 $x\in [1,n]$。请你分别找出 $A,B$ 的子集，使得它们中的元素之和相等。

$n\le {10}^6$。

题目描述

You are given two multisets $A$ and $B$ . Each multiset has exactly $n$ integers each between $1$ and $n$ inclusive. Multisets may contain multiple copies of the same number.

You would like to find a nonempty subset of $A$ and a nonempty subset of $B$ such that the sum of elements in these subsets are equal. Subsets are also multisets, i.e. they can contain elements with equal values.

If no solution exists, print $-1$ . Otherwise, print the indices of elements in any such subsets of $A$ and $B$ that have the same sum.

输入格式

The first line of the input contains a single integer $n$ ( $1<=n<=1000000$ ) — the size of both multisets.

The second line contains $n$ integers, denoting the elements of $A$ . Each element will be between $1$ and $n$ inclusive.

The third line contains $n$ integers, denoting the elements of $B$ . Each element will be between $1$ and $n$ inclusive.

输出格式

If there is no solution, print a single integer $-1$ . Otherwise, your solution should be printed on four lines.

The first line should contain a single integer $k_a$ , the size of the corresponding subset of $A$ . The second line should contain $k_a$ distinct integers, the indices of the subset of $A$ .

The third line should contain a single integer $k_b$ , the size of the corresponding subset of $B$ . The fourth line should contain $k_b$ distinct integers, the indices of the subset of $B$ .

Elements in both sets are numbered from $1$ to $n$ . If there are multiple possible solutions, print any of them.

样例 #1

样例输入 #1

10
10 10 10 10 10 10 10 10 10 10
10 9 8 7 6 5 4 3 2 1

样例输出 #1

1
2
3
5 8 10

样例 #2

样例输入 #2

5
4 4 3 3 3
2 2 2 2 5

样例输出 #2

2
2 3
2
3 5

Solution

一道思维题，需要有比较强的想象力才可能想到正解。

我们先假设这个问题一定存在一组符合题意的解，而且这组解在两个序列中还是连续的。定义 $a$ 数组的前缀和为 $suma$，$b$ 数组的前缀和为 $sumb$。定义 $c_i$ 为最靠右的下标并且满足 $sum{a}_i\ge sum{b}_{c_i}$。根据定义有：

$$sum{a}_i<sum{b}_{c_i+1}$$

$$sum{a}_i<sum{b}_{c_i}+b_{c_i+1}$$

移项得：

$$sum{a}_i-sum{b}_{c_i}<b_{c_i+1}$$

因为 $b_{c_i+1}$ 的值域是 $[1,n]$ 的，因此 $0\le sum{a}_i-sum{b}_{c_i}<n$，总共有 $n$ 个可能的取值，而 $i$ 的值域是 $[0,n]$ 的，也就是说有 $n+1$ 种可能的情况，那么根据抽屉原理，一定会存在有一组满足：

$$sum{a}_i-sum{b}_{c_i}=sum{a}_j-sum{b}_{c_j}$$

移项得：

$$sum{a}_i-sum{a}_j=sum{b}_{c_i}-sum{b}_{c_j}$$

这不就裸的前缀和式子？因此可以证明一定存在答案，并且答案一定是在两个序列中连续的一段子序列。

考虑如何求出 $c$。不难发现，因为 $suma$ 是递增的，因此 $c$ 也一定是单调不降的，所以用一个指针来计算 $c$ 即可，同时计算此时的 $sum{a}_i-sum{b}_{c_i}$，如果这一个值之前出现过，那么就是找到了一组可行的答案，否则标记这个值，继续往后找。

Code

需要注意因为序列中的数字值域是 $[1,1\times {10}^6]$ 的，并且有 $1\times {10}^6$ 个数，所以前缀和最大可以到 $1\times {10}^{12}$，所以记得开 long long

#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof a)
//#define int long long
using namespace std;
template<typename T> void read(T &k)
{
	k=0;T flag=1;char b=getchar();
	while (!isdigit(b)) {flag=(b=='-')?-1:1;b=getchar();}
	while (isdigit(b)) {k=k*10+b-48;b=getchar();}
	k*=flag;
}
template<typename T> void write(T k) {if (k<0) {putchar('-'),write(-k);return;}if (k>9) write(k/10);putchar(k%10+48);}
template<typename T> void writewith(T k,char c) {write(k);putchar(c);}
const int _SIZE=1e6;
int n,a[_SIZE+5],b[_SIZE+5];
long long suma[_SIZE+5],sumb[_SIZE+5],id[_SIZE+5][2];
bool flag[_SIZE+5];
signed main()
{
	read(n);
	for (int i=1;i<=n;i++) read(a[i]),suma[i]=suma[i-1]+a[i];
	for (int i=1;i<=n;i++) read(b[i]),sumb[i]=sumb[i-1]+b[i];
	bool swaped=0;
	if (suma[n]>sumb[n]) swaped=1,swap(suma,sumb);
	int j=0,al,ar,bl,br;
	for (int i=0;i<=n;i++)
	{
		while (suma[i]>=sumb[j] && j<=n) j++;j--;
		if (flag[suma[i]-sumb[j]])
		{
			al=id[suma[i]-sumb[j]][0]+1;ar=i;
			bl=id[suma[i]-sumb[j]][1]+1;br=j;
			break;
		}
		flag[suma[i]-sumb[j]]=1;
		id[suma[i]-sumb[j]][0]=i;
		id[suma[i]-sumb[j]][1]=j;
	}
	if (swaped) swap(al,bl),swap(ar,br);
	writewith(ar-al+1,'\n');
	for (int i=al;i<=ar;i++) writewith(i,' ');puts("");
	writewith(br-bl+1,'\n');
	for (int i=bl;i<=br;i++) writewith(i,' ');puts("");
	return 0;
}