牛客练习赛43D Tachibana Kanade Loves Sequence

题目链接：https://ac.nowcoder.com/acm/contest/548/D

题目大意

　　略

分析

　　贪心，首先小于等于 1 的数肯定不会被选到，因为选择一个数的代价是 1，必须选择大于1的数答案才有可能增加。

　　其次，答案一定是在某一个数组所有大于 1 的元素全部选完的情况下得出来的，可以用反证法证明。

　　对于 a, b 两个数组，我们可以考虑先全部选中所有大于 1 的元素，然后从那个和比较大的数组中不断扣掉数，在两数组和的大小关系不变的情况下极限扣数，扣完答案就出来了，扣数也是贪心，从小的数开始扣才能扣最多。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< string, int > PSI;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int n, a[maxN], b[maxN], c[maxN], d[maxN];
LL sumA, sumB; // 保存对应a[i]*c[i] , b[i]*d[i]的累加和 
int cntC, cntD; // 保存 c, d中1的个数 

int main(){
    INIT(); 
    cin >> n;
    Rep(i, n) {
        cin >> a[i];
        // 先把大于1的全部选中，且只有大于1的才有必要选 
        if(a[i] > 1) {
            c[i] = 1;
            ++cntC;
            sumA += a[i]; 
        }
    }
    Rep(i, n) {
        cin >> b[i];
        if(b[i] > 1) {
            d[i] = 1;
            ++cntD;
            sumB += b[i]; 
        }
    }
    
    // A指向和比较大的序列数组
    // B指向对应构造的序列数组
    // cntB指向记录B中有多少个1的变量 
    int *A, *B;
    int* cntB;
    // 最小堆 
    priority_queue< PII, vector< PII >, greater< PII > > pQ;
    LL target = abs(sumA - sumB);
    if(sumA > sumB) {
        A = a;
        B = c;
        cntB = &cntC;
    }
    else {
        A = b;
        B = d;
        cntB = &cntD;
    }
    // 排序太烦了，直接放堆里 
    Rep(i, n) if(B[i] == 1) pQ.push(MP(A[i], i));

    // 从小到大取，看看最多能从和比较大的数组中取走多少个 
    while(!pQ.empty()) {
        PII tmp = pQ.top();
        pQ.pop();
        
        target -= tmp.ft;
        if(target >= 0) {
            B[tmp.sd] = 0;
            --*cntB;
        }
        else break;
    }
    
    cout << min(sumA, sumB) - cntC - cntD << endl;
    Rep(i, n) cout << c[i] << " ";
    cout << endl;
    Rep(i, n) cout << d[i] << " ";
    cout << endl;
    return 0;
}