Wunder Fund Round 2016 F. Double Knapsack（two pointers + 鸽笼原理）

题意:

给定2个N≤106元素的multiset,元素取值范围为1∼N
现在各从中选出一些元素的subset,subset也是multiset,使得他们的和相等
有解输出各自的大小以及下标,无解输出−1

分析:

显然需要O(n)的算法,根据2个序列的prefix sum是递增的,我们可以用two pointers
假设suman≤sumbn,对于每个sumai,我们用two pointers找到最大的j使得sumai≥sumbj
获得dif=sumai−sumbj,dif显然∈[0,n−1],这样的差有n个,加上空集的suma0−sumb0=0就有(n+1)个
根据鸽笼原理必然有2个差相等,那么我们就得到了构造解

代码:

////  Created by TaoSama on 2016-01-31//  Copyright (c) 2015 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n, a[N], b[N];int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d", &n) == 1) {        LL s1 = 0, s2 = 0;        for(int i = 1; i <= n; ++i) scanf("%d", a + i), s1 += a[i];        for(int i = 1; i <= n; ++i) scanf("%d", b + i), s2 += b[i];        bool flipped = false;        if(s1 > s2) swap(a, b), flipped = true;        vector<pair<int, int> > pos(n);        for(auto &x : pos) x = { -1, -1};        pos[0] = {1, 1};        vector<int> ans1, ans2;        LL xs = 0, ys = 0;        for(int i = 1, j = 1; i <= n;) {            xs += a[i++];            while(j <= n && ys + b[j] <= xs) ys += b[j++];            int dif = xs - ys;            if(~pos[dif].first) {                for(int k = pos[dif].first; k < i; ++k) ans1.push_back(k);                for(int k = pos[dif].second; k < j; ++k) ans2.push_back(k);                break;            }            pos[dif] = {i, j};        }        if(flipped) swap(ans1, ans2);        printf("%d\n", ans1.size());        for(int i = 0; i < ans1.size(); ++i)            printf("%d%c", ans1[i], " \n"[i == ans1.size() - 1]);        printf("%d\n", ans2.size());        for(int i = 0; i < ans2.size(); ++i)            printf("%d%c", ans2[i], " \n"[i == ans2.size() - 1]);    }    return 0;}