牛客可见山峰对数量（进阶）

傷城~ 2023-06-04 14:51 2阅读 0赞

题目链接：[https://www.nowcoder.com/practice/16d1047e9fa54cea8b5170b156d89e38?tpId=101&tqId=33173&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking][https_www.nowcoder.com_practice_16d1047e9fa54cea8b5170b156d89e38_tpId_101_tqId_33173_tPage_1_rp_1_ru_ta_programmer-code-interview-guide_qru_ta_programmer-code-interview-guide_question-ranking]

### 题目大意 ###

略。

### 分析 ###

由于有重复值，每个点的贡献就不确定了，我们需要借助单调栈，因为单调栈能找到离当前山峰最近大的山峰。

具体见左神的书。

### 代码如下 ###

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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 v;
     83     istringstream sin(s);
     84     sin >> v;
     85     return v;
     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; // 最高山峰（不止一座）的高度及其位置 
    150 stack< Record > sk; // 单调栈 
    151 LL ans; 
    152 
    153 int main(){
    154     //freopen("MyOutput.txt","w",stdout);
    155     //freopen("input.txt","r",stdin);
    156     //INIT();
    157     scanf("%d", &N);
    158     Rep(i, N) {
    159         scanf("%d", &arr[i]);
    160         if(maxH < arr[i]) {
    161             maxH = arr[i];
    162             S = i;
    163         } 
    164     } 
    165     sk.push(Record(arr[S])); // 以最大值作为左边界 
    166     
    167     For(i, 1, N - 1) {
    168         int id = (S + i) % N;
    169         
    170         while(sk.top().value < arr[id]) {
    171             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 }

转载于:https://www.cnblogs.com/zaq19970105/p/11345358.html

[https_www.nowcoder.com_practice_16d1047e9fa54cea8b5170b156d89e38_tpId_101_tqId_33173_tPage_1_rp_1_ru_ta_programmer-code-interview-guide_qru_ta_programmer-code-interview-guide_question-ranking]: https://www.nowcoder.com/practice/16d1047e9fa54cea8b5170b156d89e38?tpId=101&tqId=33173&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking
[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
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