牛客最大值减去最小值小于或等于 num 的子数组数量

深碍√TFBOYSˉ_ 2023-06-04 14:51 1阅读 0赞

题目链接：[https://www.nowcoder.com/practice/5fe02eb175974e18b9a546812a17428e?tpId=101&tqId=33086&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking][https_www.nowcoder.com_practice_5fe02eb175974e18b9a546812a17428e_tpId_101_tqId_33086_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 ###

利用滑动窗口和两个双端队列维护窗口内最大值和最小值，以每个窗口左边界为基准，看它的窗口右边界最多能到多远，窗口长度即为窗口右边界的贡献。

### 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 int N, num, arr[maxN];
    140 LL ans;
    141 DQI minDQ, maxDQ;
    142 int p1 = 1, p2 = 1;
    143 
    144 int main(){
    145     //freopen("MyOutput.txt","w",stdout);
    146     //freopen("input.txt","r",stdin);
    147     //INIT();
    148     scanf("%d%d", &N, &num);
    149     For(i, 1, N) scanf("%d", &arr[i]);
    150     
    151     while(p1 <= N && p2 <= N) {
    152         while(!maxDQ.empty() && arr[maxDQ.back()] <= arr[p2]) maxDQ.pop_back();
    153         maxDQ.PB(p2);
    154         while(!minDQ.empty() && arr[minDQ.back()] >= arr[p2]) minDQ.pop_back();
    155         minDQ.PB(p2);
    156         
    157         if(arr[maxDQ.front()] - arr[minDQ.front()] <= num) ++p2;
    158         else {
    159             if(maxDQ.front() == p1) maxDQ.pop_front();
    160             if(minDQ.front() == p1) minDQ.pop_front();
    161             ans += p2 - p1;
    162             ++p1;
    163         }
    164     }
    165     while(p1 < p2) {
    166         ans += p2 - p1;
    167         ++p1;
    168     }
    169     
    170     printf("%lld\n", ans);
    171     return 0;
    172 }

转载于:https://www.cnblogs.com/zaq19970105/p/11344289.html

[https_www.nowcoder.com_practice_5fe02eb175974e18b9a546812a17428e_tpId_101_tqId_33086_tPage_1_rp_1_ru_ta_programmer-code-interview-guide_qru_ta_programmer-code-interview-guide_question-ranking]: https://www.nowcoder.com/practice/5fe02eb175974e18b9a546812a17428e?tpId=101&tqId=33086&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking
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