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 << "]" << "\n";
     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, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , 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-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;
 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] > ) {
             c[i] = ;
             ++cntC;
             sumA += a[i];
         }
     }
     Rep(i, n) {
         cin >> b[i];
         if(b[i] > ) {
             d[i] = ;
             ++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] == ) 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 >= ) {
             B[tmp.sd] = ;
             --*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 ;
 }