【LOJ】 #2132. 「NOI2015」荷马史诗

题解

k叉哈夫曼树，但是没有了二叉那样的最后一定能合并成一个树根的优秀性质，我们就不断模拟操作看看到了哪一步能用的节点数< k，然后先拿这些节点数合并起来

然后就可以k个k个合并了，大小一样先拿深度小的

代码

#include <bits/stdc++.h>
//#define ivorysi
#define enter putchar('\n')
#define space putchar(' ')
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eps 1e-8
#define mo 974711
#define MAXN 100005
#define pii pair<int,int>
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,K;
int64 W[MAXN],ql,qr,Q[MAXN],l1,ans,dep[MAXN];
void Solve() {
    read(N);read(K);
    for(int i = 1 ; i <= N ; ++i) read(W[i]);
    sort(W + 1,W + N + 1);
    l1 = 1;ql = 1,qr = 0;
    int t = N;
    while(t >= K) t = t - K + 1;
    if(t != 1) {
	for(int i = 1 ; i <= t ; ++i) {
	    ans += W[i];
	    ++l1;
	}
	Q[++qr] = ans;
	dep[qr] = 1;
    }
    t = N - t + 1;
    while(t != 1) {
	int64 tmp = 0,d = 1;
	for(int j = 1 ; j <= K ; ++j) {
	    int64 t;
	    if(ql <= qr && l1 <= N) {
		if(W[l1] <= Q[ql]) t = W[l1],++l1;
		else t = Q[ql],d = max(d,dep[ql] + 1),++ql;
	    }
	    else if(ql <= qr) {
		t = Q[ql],d = max(d,dep[ql] + 1),++ql;
	    }
	    else t = W[l1],++l1;
	    tmp += t;
	}
	ans += tmp;
	Q[++qr] = tmp;
	dep[qr] = d;
	t = t - K + 1;
    }
    out(ans);enter;out(dep[qr]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}