#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#include <bits/stdc++.h>
#include <immintrin.h>
using namespace std;
const int PREFETCH_DIST = 64;
// 辅助函数:向地址 p 写入 3 字节
inline void store3(uint8_t* __restrict__ p, uint32_t val) {
*(uint32_t*)p = val;
}
// 辅助函数:向地址 p 写入 2 字节
inline void store2(uint8_t* __restrict__ p, uint16_t val) {
*(uint16_t*)p = val;
}
// 辅助函数:向地址 p 写入 1 字节
inline void store1(uint8_t* __restrict__ p, uint8_t val) {
*p = val;
}
void sort(uint* a, int n) {
// ---------------------------------------------------------
// 内存规划:
// a: 输入/最终输出 (4N)。Pass 1 之后数据移至 b,a 用作 Pass 2 的输出 (2N) 和 Pass 4 的最终输出。
// b: 临时 buffer (3N + Padding)。用于 Pass 1 输出,并在 Pass 3 复用。
// ---------------------------------------------------------
uint8_t* b = (uint8_t*)malloc((size_t)n * 3 + 256 * 64 + 4096);
uint cnt_global[256];
memset(cnt_global, 0, sizeof(cnt_global));
// 1.1 统计 B3 (Global MSD)
for (int i = 0; i < n; i++) {
cnt_global[a[i] >> 24]++;
}
// 1.2 计算 B3 Offset
uint ptr_global[256];
size_t offset_b3 = 0;
for (int i = 0; i < 256; i++) {
ptr_global[i] = (uint)offset_b3;
offset_b3 += cnt_global[i] * 3 + 64; // Padding needed for store3/prefetch
}
// ---------------------------------------------------------
// Pass 1: Global MSD (Partition by B3)
// Read: a (4B) -> Write: b (3B: [B0, B1, B2])
// ---------------------------------------------------------
{
uint* __restrict__ src = a;
uint8_t* __restrict__ dst = b;
uint p[256];
memcpy(p, ptr_global, sizeof(p));
for (int i = 0; i < n; i += 16) {
_mm_prefetch((const char*)&src[i + PREFETCH_DIST], _MM_HINT_NTA);
#pragma GCC unroll 16
for (int j = 0; j < 16; j++) {
uint val = src[i + j];
uint8_t k = val >> 24;
store3(dst + p[k], val);
p[k] += 3;
}
}
}
// ---------------------------------------------------------
// 分段处理:遍历 B3 的每一个 Bucket
// ---------------------------------------------------------
uint8_t* a_u8 = (uint8_t*)a;
size_t a_offset_base = 0; // 当前 B3 Bucket 在 a 中的起始元素下标
for (int i_b3 = 0; i_b3 < 256; i_b3++) {
int count_b3 = cnt_global[i_b3];
if (count_b3 == 0) continue;
uint8_t* seg_b_in = b + ptr_global[i_b3];
// Pass 2 输出到 a 的当前段。
// Pass 2 写 2 字节/元素,Pass 4 写 4 字节/元素。
// 由于是顺序处理,Pass 4 的写入会覆盖 Pass 2 的数据,但 Pass 3 会在此之前将 Pass 2 数据搬回 b。
uint8_t* seg_a_current = a_u8 + a_offset_base * 4;
// -----------------------------------------------------
// Pass 2: Local MSD (Partition by B2)
// Read b (3B: [B0, B1, B2]) -> Write a (2B: [B0, B1])
// -----------------------------------------------------
uint cnt_b2[256] = {0};
// 统计 B2 (Offset 2 in b)
for (int i = 0; i < count_b3; i++) {
cnt_b2[*(seg_b_in + i * 3 + 2)]++;
}
uint ptr_b2[256];
uint32_t tmp_offset = 0;
// 计算 Pass 2 写入偏移。注意 a 必须紧凑 (No Padding),因为它是最终输出容器。
for (int k = 0; k < 256; k++) {
ptr_b2[k] = tmp_offset;
tmp_offset += cnt_b2[k] * 4;
}
// 保存 B2 的信息供 Pass 3/4 使用
uint cnt_b2_saved[256];
memcpy(cnt_b2_saved, cnt_b2, sizeof(cnt_b2));
uint ptr_b2_saved[256];
memcpy(ptr_b2_saved, ptr_b2, sizeof(ptr_b2));
// 执行 Pass 2 Scatter
{
uint p[256];
memcpy(p, ptr_b2, sizeof(p));
uint8_t* src = seg_b_in;
uint8_t* dst = seg_a_current;
int i = 0;
for (; i <= count_b3 - 16; i += 16) {
_mm_prefetch((const char*)&src[(i + PREFETCH_DIST) * 3], _MM_HINT_NTA);
#pragma GCC unroll 16
for (int j = 0; j < 16; j++) {
uint32_t val = *(uint32_t*)(src + (i + j) * 3);
uint8_t key = (val >> 16) & 0xFF; // B2
store2(dst + p[key], val); // Store [B0, B1]
p[key] += 2;
}
}
for (; i < count_b3; i++) {
uint32_t val = *(uint32_t*)(src + i * 3);
uint8_t key = (val >> 16) & 0xFF;
store2(dst + p[key], val);
p[key] += 2;
}
}
// -----------------------------------------------------
// 分段处理:遍历 B2 的每一个 Bucket
// -----------------------------------------------------
for (int i_b2 = 0; i_b2 < 256; i_b2++) {
int count_b2 = cnt_b2_saved[i_b2];
if (count_b2 == 0) continue;
uint8_t* seg_a_in = seg_a_current + ptr_b2_saved[i_b2];
// 复用 b 的空间作为 Pass 3 的输出 buffer。
// 此时 b 对应的 B3 Bucket 区域已读完,可以复用。
// Pass 3 需要 1N 空间,b 有 3N 空间,足够。
uint8_t* seg_b_temp = seg_b_in;
// 同时统计 B0 (Pass 3 Key) 和 B1 (Pass 4 Key)
uint cnt_b0[256] = {0};
uint cnt_b1[256] = {0};
for (int i = 0; i < count_b2; i++) {
uint16_t val = *(uint16_t*)(seg_a_in + i * 2);
cnt_b0[val & 0xFF]++;
cnt_b1[val >> 8]++;
}
// 计算 Pass 3 (Write b) 指针 (Need Padding)
uint ptr_b0[256];
tmp_offset = 0;
for (int k = 0; k < 256; k++) {
ptr_b0[k] = tmp_offset;
tmp_offset += cnt_b0[k] + 32;
}
uint ptr_b0_read[256]; // 用于 Pass 4 读取
memcpy(ptr_b0_read, ptr_b0, sizeof(ptr_b0));
// 计算 Pass 4 (Write a Final) 指针 (No Padding)
// 目标地址是 a 中对应 B2 Bucket 的位置
size_t elem_offset_b2 = ptr_b2_saved[i_b2] / 2;
uint* a_final_dst_base = (uint*)a_u8 + a_offset_base + elem_offset_b2;
uint ptr_b1[256];
tmp_offset = 0;
for (int k = 0; k < 256; k++) {
ptr_b1[k] = tmp_offset;
tmp_offset += cnt_b1[k];
}
// -------------------------------------------------
// Pass 3: LSD Step 1 (Key B0)
// Read a (2B: [B0, B1]) -> Write b (1B: [B1])
// -------------------------------------------------
{
uint p[256];
memcpy(p, ptr_b0, sizeof(p));
uint8_t* src = seg_a_in;
uint8_t* dst = seg_b_temp;
int i = 0;
for (; i <= count_b2 - 16; i += 16) {
_mm_prefetch((const char*)&src[(i + PREFETCH_DIST) * 2], _MM_HINT_NTA);
#pragma GCC unroll 16
for (int j = 0; j < 16; j++) {
uint16_t val = *(uint16_t*)(src + (i + j) * 2);
uint8_t key = val & 0xFF; // B0
store1(dst + p[key], val >> 8); // Store B1
p[key]++;
}
}
for (; i < count_b2; i++) {
uint16_t val = *(uint16_t*)(src + i * 2);
uint8_t key = val & 0xFF;
store1(dst + p[key], val >> 8);
p[key]++;
}
}
// -------------------------------------------------
// Pass 4: LSD Step 2 (Key B1) & Finalize
// Read b (1B: [B1]) -> Write a (4B: Full)
// -------------------------------------------------
{
uint p[256];
memcpy(p, ptr_b1, sizeof(p));
uint32_t high_bits = (i_b3 << 24) | (i_b2 << 16);
// 按照 B0 的顺序遍历 (Pass 3 的输出桶)
// 这样进入 Pass 4 Scatter 的数据流是按 B0 排序的
// Pass 4 是稳定 Scatter,所以最终结果按 B1 排序,且相同 B1 内按 B0 排序
for (int b0 = 0; b0 < 256; b0++) {
int count = cnt_b0[b0];
if (count == 0) continue;
uint8_t* src = seg_b_temp + ptr_b0_read[b0];
// 构建除 B1 外的部分
uint32_t val_base = high_bits | b0;
int i = 0;
for (; i <= count - 16; i += 16) {
#pragma GCC unroll 16
for (int j = 0; j < 16; j++) {
uint8_t b1 = src[i + j]; // Read B1
// Write Full: B3 | B2 | B1 | B0
a_final_dst_base[p[b1]] = val_base | (b1 << 8);
p[b1]++;
}
}
for (; i < count; i++) {
uint8_t b1 = src[i];
a_final_dst_base[p[b1]] = val_base | (b1 << 8);
p[b1]++;
}
}
}
}
a_offset_base += count_b3;
}
free(b);
}
| Compilation | N/A | N/A | Compile OK | Score: N/A | 显示更多 |
| Testcase #1 | 786.266 ms | 667 MB + 648 KB | Runtime Error | Score: 0 | 显示更多 |