CF618F Double Knapsack

原题链接：http://codeforces.com/problemset/problem/618/F

Double Knapsack

题目描述

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

You would like to find a nonempty subset of A A and a nonempty subset of B 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 −1 . Otherwise, print the indices of elements in any such subsets of A A and B B that have the same sum.

输入输出格式

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 n integers, denoting the elements of A . Each element will be between 1 1 and n n inclusive.

The third line contains n 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 −1 . Otherwise, your solution should be printed on four lines.

The first line should contain a single integer kakak_{a} 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 A .

The third line should contain a single integer kbkbk_{b} , the size of the corresponding subset of B B . The fourth line should contain kbkbk_{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.

输入输出样例

输入样例#1：

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

输出样例#1：

1
2
3
5 8 10

输入样例#2：

5
4 4 3 3 3
2 2 2 2 5

输出样例#2：

2
2 3
2
3 5

题目大意

给你两个可重集 A,B ,A、B的元素个数都为n且n⩽1000000，它们中每个元素的大小x∈[1,n]。请你分别找出A ，B的可重子集，使得它们中的元素之和相等。

输入格式：

第一行为一个整数n，表示两个子集的大小。

第二、三行皆为n个整数，分别表示AA 、BB 的元素。

输出格式：

如果无解，请输出-1。如果有解，第一行输出A 的可重子集中元素的个数，第二行输出该子集中元素在A 中对应的下标；第三行输出B 的可重子集中元素的个数，第四行输出该子集中元素在B 中对应的下标。

数据可能存在多组解，输出一组即可。

题解

我们对于两个数列分别求前缀和，设两个前缀和分别为f(a)=∑ai=1Ai,g(b)=∑bi=1Bif(a)=∑i=1aAi,g(b)=∑i=1bBif(a)=\sum _ {i=1}^{a}A_i,g(b)=\sum_ {i=1}^{b}B_i，假设f(n)<g(n)f(n)<g(n)f(n)，即AAA数列中的值小于B" role="presentation" style="position: relative;">BBB数列。找一个最小的g(b)g(b)g(b)使得g(b)⩾f(a)g(b)⩾f(a)g(b)\geqslant f(a)，那么肯定有g(b)−f(a)∈[0,n−1]g(b)−f(a)∈[0,n−1]g(b)-f(a)\in [0,n-1]，则共有nnn种取值，而因为我们从0开始求解，会求出n+1" role="presentation" style="position: relative;">n+1n+1n+1个取值，所以肯定会有两个重复的取值，那么这两个前缀之间的串的和肯定是一样的。

代码

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int M=1e6+5;
struct sd{int cot,t1,t2;
};
int n;
ll x1[M],x2[M];
sd bak[M];
void in()
{scanf("%lld",&n);for(int i=1;i<=n;++i)scanf("%lld",&x1[i]),x1[i]+=x1[i-1];for(int i=1;i<=n;++i)scanf("%lld",&x2[i]),x2[i]+=x2[i-1];
}
void ac()
{bool flag=0;if(x1[n]>x2[n])flag=1,swap(x1,x2);int j=0,hh,s1=1,s2=1,e1=1,e2=1;for(int i=0;i<=n;++i){while(x2[j]<x1[i])++j;hh=x2[j]-x1[i];bak[hh].cot++;if(bak[hh].cot>=2){s1=bak[hh].t1;s2=bak[hh].t2;e1=i;e2=j;break;}bak[hh].t1=i;bak[hh].t2=j;}if(flag)swap(x1,x2),swap(e1,e2),swap(s1,s2);printf("%d\n",e1-s1);for(int i=s1+1;i<=e1;++i)printf("%d ",i);printf("\n%d\n",e2-s2);for(int i=s2+1;i<=e2;++i)printf("%d ",i);
}
int main()
{   in();ac();return 0;
}