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

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "
";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< string, int > PSI;
 69 typedef set< int > SI;
 70 typedef vector< int > VI;
 71 typedef map< int, int > MII;
 72 typedef pair< LL, LL > PLL;
 73 typedef vector< LL > VL;
 74 typedef vector< VL > VVL;
 75 const double EPS = 1e-10;
 76 const LL inf = 0x7fffffff;
 77 const LL infLL = 0x7fffffffffffffffLL;
 78 const LL mod = 1e9 + 7;
 79 const int maxN = 1e5 + 7;
 80 const LL ONE = 1;
 81 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 82 const LL oddBits = 0x5555555555555555;
 83 
 84 int n, a[maxN], b[maxN], c[maxN], d[maxN];
 85 LL sumA, sumB; // 保存对应a[i]*c[i] , b[i]*d[i]的累加和 
 86 int cntC, cntD; // 保存 c, d中1的个数 
 87 
 88 int main(){
 89     INIT(); 
 90     cin >> n;
 91     Rep(i, n) {
 92         cin >> a[i];
 93         // 先把大于1的全部选中，且只有大于1的才有必要选 
 94         if(a[i] > 1) {
 95             c[i] = 1;
 96             ++cntC;
 97             sumA += a[i]; 
 98         }
 99     }
100     Rep(i, n) {
101         cin >> b[i];
102         if(b[i] > 1) {
103             d[i] = 1;
104             ++cntD;
105             sumB += b[i]; 
106         }
107     }
108     
109     // A指向和比较大的序列数组
110     // B指向对应构造的序列数组
111     // cntB指向记录B中有多少个1的变量 
112     int *A, *B;
113     int* cntB;
114     // 最小堆 
115     priority_queue< PII, vector< PII >, greater< PII > > pQ;
116     LL target = abs(sumA - sumB);
117     if(sumA > sumB) {
118         A = a;
119         B = c;
120         cntB = &cntC;
121     }
122     else {
123         A = b;
124         B = d;
125         cntB = &cntD;
126     }
127     // 排序太烦了，直接放堆里 
128     Rep(i, n) if(B[i] == 1) pQ.push(MP(A[i], i));
129 
130     // 从小到大取，看看最多能从和比较大的数组中取走多少个 
131     while(!pQ.empty()) {
132         PII tmp = pQ.top();
133         pQ.pop();
134         
135         target -= tmp.ft;
136         if(target >= 0) {
137             B[tmp.sd] = 0;
138             --*cntB;
139         }
140         else break;
141     }
142     
143     cout << min(sumA, sumB) - cntC - cntD << endl;
144     Rep(i, n) cout << c[i] << " ";
145     cout << endl;
146     Rep(i, n) cout << d[i] << " ";
147     cout << endl;
148     return 0;
149 }

View Code