The number of cattle off the mountain visible (Advanced)

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Subject to the effect

  slightly.

analysis

  Because duplicate values, the contribution of each point not identified, we need the help of monotonous stack, the stack can be found away from the monotony because the current peaks nearest big mountain.
  See in particular left a book of God.

code show as below

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
  6 #define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
  7 #define rFor(i,t,s) for (int i = (int)(t); i >= (int)(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 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
 21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
 22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
 23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 24  
 25 #define ms0(a) memset(a,0,sizeof(a))
 26 #define msI(a) memset(a,0x3f,sizeof(a))
 27 #define msM(a) memset(a,-1,sizeof(a))
 28 
 29 #define MP make_pair
 30 #define PB push_back
 31 #define ft first
 32 #define sd second
 33  
 34 template<typename T1, typename T2>
 35 istream &operator>>(istream &in, pair<T1, T2> &p) {
 36     in >> p.first >> p.second;
 37     return in;
 38 }
 39  
 40 template<typename T>
 41 istream &operator>>(istream &in, vector<T> &v) {
 42     for (auto &x: v)
 43         in >> x;
 44     return in;
 45 }
 46 
 47 template<typename T>
 48 ostream &operator<<(ostream &out, vector<T> &v) {
 49     Rep(i, v.size()) out << v[i] << " \n"[i == v.size()];
 50     return out;
 51 }
 52  
 53 template<typename T1, typename T2>
 54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 55     out << "[" << p.first << ", " << p.second << "]" << "\n";
 56     return out;
 57 }
 58 
 59 inline int gc(){
 60     static const int BUF = 1e7;
 61     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 62     
 63     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 64     return *bg++;
 65 } 
 66 
 67 inline int ri(){
 68     int x = 0, f = 1, c = gc();
 69     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 70     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 71     return x*f;
 72 }
 73 
 74 template<class T>
75 inline String toString (T x) {
 76      ostringstream Sout;
77      Sout << x;
78      return sout.str ();
79  }
 80  
81 inline int toInt ( string s) {
 82      int in;
83      istringstream son (s);
84      son >> in;
85      return to;
86  }
 87  
88  // min <= AIM <= max 
89 template <typename T>
 90 inline bool BETWEEN(const T aim, const T min, const T max) {
 91     return min <= aim && aim <= max;
 92 }
 93  
 94 typedef long long LL;
 95 typedef unsigned long long uLL;
 96 typedef vector< int > VI;
 97 typedef vector< bool > VB;
 98 typedef vector< char > VC;
 99 typedef vector< double > VD;
100 typedef vector< string > VS;
101 typedef vector< LL > VL;
102 typedef vector< VI > VVI;
103 typedef vector< VB > VVB;
104 typedef vector< VS > VVS;
105 typedef vector< VL > VVL;
106 typedef vector< VVI > VVVI;
107 typedef vector< VVL > VVVL;
108 typedef pair< int, int > PII;
109 typedef pair< LL, LL > PLL;
110 typedef pair< int, string > PIS;
111 typedef pair< string, int > PSI;
112 typedef pair< string, string > PSS;
113 typedef pair< double, double > PDD;
114 typedef vector< PII > VPII;
115 typedef vector< PLL > VPLL;
116 typedef vector< VPII > VVPII;
117 typedef vector< VPLL > VVPLL;
118 typedef vector< VS > VVS;
119 typedef map< int, int > MII;
120 typedef unordered_map< int, int > uMII;
121 typedef map< LL, LL > MLL;
122 typedef map< string, int > MSI;
123 typedef map< int, string > MIS;
124 typedef set< int > SI;
125 typedef stack< int > SKI;
126 typedef deque< int > DQI;
127 typedef queue< int > QI;
128 typedef priority_queue< int > PQIMax;
129 typedef priority_queue< int, VI, greater< int > > PQIMin;
130 const double EPS = 1e-8;
131 const LL inf = 0x7fffffff;
132 const LL infLL = 0x7fffffffffffffffLL;
133 const LL mod = 1e9 + 7;
134 const int maxN = 1e6 + 7;
135 const LL ONE = 1;
136 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
137 const LL oddBits = 0x5555555555555555;
138 
139 struct Record {
140     int value, times;
141         
142     Record(int v) {
143         value = v;
144         times = 1;
145     }
146 };
147 
148 int N, arr[maxN];
149 int maxH = -inf, S; // the highest peak (more than one) and the height position of 
150 Stack <the Record> SK; // monotone stack 
151  LL ANS; 
 152  
153  int main () {
 154      // The freopen ( "MyOutput.txt", "W", stdout);
 155      // The freopen ( "input.txt", "R & lt", stdin);
 156      // the INIT (); 
157      Scanf ( " % D " , & N);
 158      the Rep (I, N) {
 159          Scanf ( " % D " , & ARR [I]);
 160.          IF (maxH < arr[i]) {
161             maxH =ARR [I];
 162              S = I;
 163          } 
 164 is      } 
 165      sk.push (the Record (ARR [S])); // the maximum value of as the left boundary 
166      
167 is      the For (I, . 1 , N - . 1 ) {
 168          int ID = (I + S)% N;
 169          
170.          the while (. sk.top () value < ARR [ID]) {
 171 is              int tmp = sk.top () Times;. sk.pop ();
 172              
173              ANS = + (tmp - . 1 ) * tmp / 2 + 2 * tmp;
 174         }
175         if(sk.top().value == arr[id]) ++sk.top().times;
176         else sk.push(Record(arr[id]));
177     }
178     while(!sk.empty()) {
179         int tmp = sk.top().times; sk.pop();
180         
181         ans += (tmp - 1) * tmp / 2; // 内部产生山峰对 
182         
183         if(sk.size() == 1 && sk.top().times == 1) ans += tmp;
184         else if(sk.size() != 0) ans += 2 * tmp;
185     }
186     
187     printf("%d\n", ans);
188     return 0;
189 }
View Code

 

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Origin www.cnblogs.com/zaq19970105/p/11345358.html