結果

問題 No.2304 Distinct Elements
ユーザー 👑 ygussany
提出日時 2023-04-29 18:32:56
言語 C
(gcc 13.3.0)
結果
AC  
実行時間 112 ms / 3,000 ms
コード長 4,041 bytes
コンパイル時間 404 ms
コンパイル使用メモリ 33,152 KB
実行使用メモリ 40,848 KB
最終ジャッジ日時 2024-11-18 12:37:02
合計ジャッジ時間 4,640 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 58
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <stdio.h>
#include <stdlib.h>
void merge_sort(int n, int x[])
{
static int y[200001] = {};
if (n <= 1) return;
merge_sort(n / 2, &(x[0]));
merge_sort((n + 1) / 2, &(x[n/2]));
int i, p, q;
for (i = 0, p = 0, q = n / 2; i < n; i++) {
if (p >= n / 2) y[i] = x[q++];
else if (q >= n) y[i] = x[p++];
else y[i] = (x[p] < x[q])? x[p++]: x[q++];
}
for (i = 0; i < n; i++) x[i] = y[i];
}
typedef struct {
int key, id, left;
} data;
typedef struct {
data obj[400001];
int size;
} max_heap;
void push(max_heap* h, data x)
{
int i = ++(h->size), j = i >> 1;
data tmp;
h->obj[i] = x;
while (j > 0) {
if (h->obj[i].key > h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
i = j;
j >>= 1;
} else break;
}
}
data pop(max_heap* h)
{
int i = 1, j = 2;
data output = h->obj[1], tmp;
h->obj[1] = h->obj[(h->size)--];
while (j <= h->size) {
if (j < h->size && h->obj[j^1].key > h->obj[j].key) j ^= 1;
if (h->obj[j].key > h->obj[i].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
i = j;
j <<= 1;
} else break;
}
return output;
}
typedef struct {
data *obj;
int size, max_size, thr;
} min_heap;
void init_min_heap(min_heap* h, int s)
{
h->max_size = s;
h->size = 0;
h->thr = 0;
h->obj = (data*)malloc(sizeof(data) * (h->max_size + 1));
}
void min_push(min_heap* h, data x)
{
int i = ++(h->size), j = i >> 1;
data tmp;
if (h->size > h->max_size) {
h->max_size <<= 1;
h->obj = (data*)realloc(h->obj, sizeof(data) * (h->max_size + 1));
}
h->obj[i] = x;
while (j > 0) {
if (h->obj[i].key < h->obj[j].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
i = j;
j >>= 1;
} else break;
}
}
data min_pop(min_heap* h)
{
int i = 1, j = 2;
data output = h->obj[1], tmp;
h->obj[1] = h->obj[(h->size)--];
while (j <= h->size) {
if (j < h->size && h->obj[j^1].key < h->obj[j].key) j ^= 1;
if (h->obj[j].key < h->obj[i].key) {
tmp = h->obj[j];
h->obj[j] = h->obj[i];
h->obj[i] = tmp;
i = j;
j <<= 1;
} else break;
}
return output;
}
void merge_min_heap(min_heap* h, min_heap* hh)
{
int i;
data d;
if (h->size >= hh->size) {
for (i = 1; i <= hh->size; i++) {
d = hh->obj[i];
d.key += h->thr - hh->thr;
min_push(h, d);
}
free(hh->obj);
} else {
for (i = 1; i <= h->size; i++) {
d = h->obj[i];
d.key += hh->thr - h->thr;
min_push(hh, d);
}
free(h->obj);
*h = *hh;
}
}
int main()
{
int i, N, A[200001];
scanf("%d", &N);
for (i = 0; i < N; i++) scanf("%d", &(A[i]));
merge_sort(N, A);
int j, k;
long long ans = 0;
max_heap h;
min_heap he[200001];
data d, e;
h.size = 0;
d.key = A[0];
d.left = A[0];
d.id = 0;
push(&h, d);
init_min_heap(&(he[0]), 10);
for (i = 1; i < N; i++) {
d = pop(&h);
if (A[i] > d.key + 1) {
push(&h, d);
d.key = A[i];
d.left = A[i];
d.id = i;
push(&h, d);
init_min_heap(&(he[i]), 10);
} else if (A[i] == d.key + 1) {
d.key++;
push(&h, d);
} else {
j = d.id;
while (he[j].size > 0 && he[j].obj[1].key <= he[j].thr) min_pop(&(he[j]));
if (abs(d.key - A[i]) - he[j].size + (d.key - d.left + 1 - he[j].size) < abs(d.key + 1 - A[i])) {
ans += abs(d.key - A[i]) - he[j].size + (d.key - d.left + 1 - he[j].size);
d.left--;
he[j].thr++;
if (h.size > 0 && h.obj[1].key == d.left - 1) {
h.obj[1].key = d.key;
merge_min_heap(&(he[h.obj[1].id]), &(he[j]));
e.key = d.key - A[i] + he[h.obj[1].id].thr;
min_push(&(he[h.obj[1].id]), e);
} else {
push(&h, d);
e.key = d.key - A[i] + he[j].thr;
min_push(&(he[j]), e);
}
} else {
ans += abs(d.key + 1 - A[i]);
d.key++;
push(&h, d);
e.key = d.key - A[i] + he[j].thr;
min_push(&(he[j]), e);
}
}
}
printf("%lld\n", ans);
fflush(stdout);
return 0;
}
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