P2048 [NOI2010]超级钢琴 [堆+st表]

考虑只能取长度为 [L,R] 的，然后不难想到用堆搞。

搞个前缀和的st表，里面维护的是一个 最大值的位置

struct rmq {
    int mx[N][20] ;
    void qwq(int n) {
        rep(i , 1 , n) mx[i][0] = i ;
        for(int j = 1 ; (1 << j) <= n ; j ++)
            for(int i = 1 ; i + (1 << j) - 1 <= n ; i ++) {
                int x = mx[i][j - 1] , y = mx[i + (1 << j - 1)][j - 1] ;
                mx[i][j] = sum[x] > sum[y] ? x : y ;
            }
    }
    int qry(int l , int r) {
        int k = log2(r - l + 1) ;
        int x = mx[l][k] , y = mx[r - (1 << k) + 1][k] ;
        return sum[x] > sum[y] ? x : y ;
    }
} st ;

考虑 对于任意的 \(i\) , 放进去一个 \([i+L-1 , min(i+R-1 , n)]\)

每次从中间分裂就好了，堆自动取max，这题没了
code

// by Isaunoya
#include 
using namespace std;

#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
#define int long long

const int _ = 1 << 21;
struct I {
    char fin[_], *p1 = fin, *p2 = fin;
    inline char gc() {
        return (p1 == p2) && (p2 = (p1 = fin) + fread(fin, 1, _, stdin), p1 == p2) ? EOF : *p1++;
    }
    inline I& operator>>(int& x) {
        bool sign = 1;
        char c = 0;
        while (c < 48) ((c = gc()) == 45) && (sign = 0);
        x = (c & 15);
        while ((c = gc()) > 47) x = (x << 1) + (x << 3) + (c & 15);
        x = sign ? x : -x;
        return *this;
    }
    inline I& operator>>(double& x) {
        bool sign = 1;
        char c = 0;
        while (c < 48) ((c = gc()) == 45) && (sign = 0);
        x = (c - 48);
        while ((c = gc()) > 47) x = x * 10 + (c - 48);
        if (c == '.') {
            double d = 1.0;
            while ((c = gc()) > 47) d = d * 0.1, x = x + (d * (c - 48));
        }
        x = sign ? x : -x;
        return *this;
    }
    inline I& operator>>(char& x) {
        do
            x = gc();
        while (isspace(x));
        return *this;
    }
    inline I& operator>>(string& s) {
        s = "";
        char c = gc();
        while (isspace(c)) c = gc();
        while (!isspace(c) && c != EOF) s += c, c = gc();
        return *this;
    }
} in;
struct O {
    char st[100], fout[_];
    signed stk = 0, top = 0;
    inline void flush() {
        fwrite(fout, 1, top, stdout), fflush(stdout), top = 0;
    }
    inline O& operator<<(int x) {
        if (top > (1 << 20)) flush();
        if (x < 0) fout[top++] = 45, x = -x;
        do
            st[++stk] = x % 10 ^ 48, x /= 10;
        while (x);
        while (stk) fout[top++] = st[stk--];
        return *this;
    }
    inline O& operator<<(char x) {
        fout[top++] = x;
        return *this;
    }
    inline O& operator<<(string s) {
        if (top > (1 << 20)) flush();
        for (char x : s) fout[top++] = x;
        return *this;
    }
} out;
#define pb emplace_back
#define fir first
#define sec second

template < class T > inline void cmax(T & x , const T & y) {
    (x < y) && (x = y) ;
}
template < class T > inline void cmin(T & x , const T & y) {
    (x > y) && (x = y) ;
}

int n , k , L , R ;
const int N = 5e5 + 10 ;
int sum[N] ;
struct rmq {
    int mx[N][20] ;
    void qwq(int n) {
        rep(i , 1 , n) mx[i][0] = i ;
        for(int j = 1 ; (1 << j) <= n ; j ++)
            for(int i = 1 ; i + (1 << j) - 1 <= n ; i ++) {
                int x = mx[i][j - 1] , y = mx[i + (1 << j - 1)][j - 1] ;
                mx[i][j] = sum[x] > sum[y] ? x : y ;
            }
    }
    int qry(int l , int r) {
        int k = log2(r - l + 1) ;
        int x = mx[l][k] , y = mx[r - (1 << k) + 1][k] ;
        return sum[x] > sum[y] ? x : y ;
    }
} st ;
struct element {
    int o , l , r , t ;
    element() {}
    element(int _o , int _l , int _r) : o(_o) , l(_l) , r(_r) , t(st.qry(_l , _r)) {}
    bool operator < (const element & other) const {
        return sum[t] - sum[o - 1] < sum[other.t] - sum[other.o - 1] ;
    }
};

priority_queue < element > q ;
signed main() {
#ifdef _WIN64
    freopen("testdata.in" , "r" , stdin) ;
#endif
    in >> n >> k >> L >> R ;
    rep(i , 1 , n) {
        in >> sum[i] ;
        sum[i] += sum[i - 1] ;
    }
    st.qwq(n) ;
    rep(i , 1 , n) 
        if(i + L - 1 <= n) q.push(element(i , i + L - 1 , min(i + R - 1 , n))) ;
    int ans = 0 ;
    rep(i , 1 , k) {
        int o = q.top().o , l = q.top().l , r = q.top().r , t = q.top().t ; q.pop() ;
        ans += sum[t] - sum[o - 1] ;
        if(l != t) q.push(element(o , l , t - 1)) ;
        if(r != t) q.push(element(o , t + 1 , r)) ;
    }
    out << ans << '\n' ;
    return out.flush(), 0;
}