我的就像插入排序。对于每一朵新花，它从后到前检查它前面的那朵是否挡住了它；如果是这样，则意味着它必须放在它后面。同样，它也会从前向后搜索并检查它后面的人是否阻止它；如果是这样，则意味着它必须放在它的前面。如果没有块，它只会检查最佳点高度。
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <stdbool.h>
#define uint32 uint32_t
static void
Swap(int *AIdx, int *BIdx)
{
int Tmp = *AIdx;
*AIdx = *BIdx;
*BIdx = Tmp;
}
static void
SwapTo(int Start, int End, int *Array)
{
while(Start != End)
{
Swap(&Array[Start], &Array[Start - 1]);
--Start;
}
}
static void
PrintTo(int End, int *Array)
{
for(int Idx = 0;
Idx < End;
++Idx)
{
printf("%d, ", Array[Idx]);
}
printf("\n");
}
/* Does A block B? */
static bool
Blocks(int AIdx, int BIdx, int *Heights, int *Blooms, int *Wilts)
{
bool Result = (Heights[AIdx] > Heights[BIdx] &&
Wilts[AIdx] >= Blooms[BIdx] &&
Blooms[AIdx] <= Wilts[BIdx]);
return Result;
}
static void
Order(int *Heights, int *Blooms, int *Wilts,
int FlowerCount)
{
for(int FlowerIdx = 1;
FlowerIdx < FlowerCount;
++FlowerIdx)
{
PrintTo(FlowerIdx, Heights);
/* front to back */
int MinIdx = -1;
for(int Idx = 0;
Idx < FlowerIdx;
++Idx)
{
if(Blocks(Idx, FlowerIdx, Heights, Blooms, Wilts))
{
MinIdx = Idx;
break;
}
}
/* back to front */
int MaxIdx = -1;
for(int Idx = (FlowerIdx - 1);
Idx >= 0;
--Idx)
{
if(Blocks(FlowerIdx, Idx, Heights, Blooms, Wilts))
{
MaxIdx = (Idx + 1);
break;
}
}
/* best height index */
int BestHeightIdx = -1;
if(MinIdx == -1 &&
MaxIdx == -1)
{
for(int Idx = 0;
Idx < FlowerIdx;
++Idx)
{
if(Heights[FlowerIdx] > Heights[Idx])
{
BestHeightIdx = Idx;
break;
}
}
if(BestHeightIdx == -1)
{
BestHeightIdx = FlowerIdx;
}
}
int SwapToIdx = -1;
if((MaxIdx == -1 && MinIdx != -1) ||
(MinIdx == -1 && MaxIdx != -1) ||
(MaxIdx != -1 && MinIdx != -1 && MaxIdx == MinIdx))
{
SwapToIdx = (MinIdx != -1) ? MinIdx : MaxIdx;
}
else if(BestHeightIdx != -1)
{
SwapToIdx = BestHeightIdx;
}
else
{
fprintf(stderr, "Spot-finding error:\n MinIdx: %d, MaxIdx: %d, BestHIdx: %d\n",
MinIdx, MaxIdx, BestHeightIdx);
exit(1);
}
SwapTo(FlowerIdx, SwapToIdx, Heights);
SwapTo(FlowerIdx, SwapToIdx, Blooms);
SwapTo(FlowerIdx, SwapToIdx, Wilts);
}
}
int
main(int argc, char *argv[])
{
int Heights0[] = {5,4,3,2,1};
int Blooms0[] = {1,1,1,1,1};
int Wilts0[] = {365,365,365,365,365};
int Heights1[] = {5,4,3,2,1};
int Blooms1[] = {1,5,10,15,20};
int Wilts1[] = {4,9,14,19,24};
int Heights2[] = {5,4,3,2,1};
int Blooms2[] = {1,5,10,15,20};
int Wilts2[] = {5,10,15,20,25};
int Heights3[] = {5,4,3,2,1};
int Blooms3[] = {1,5,10,15,20};
int Wilts3[] = {5,10,14,20,25};
int Heights4[] = {1,2,3,4,5,6};
int Blooms4[] = {1,3,1,3,1,3};
int Wilts4[] = {2,4,2,4,2,4};
int Heights5[] = {3,2,5,4};
int Blooms5[] = {1,2,11,10};
int Wilts5[] = {4,3,12,13};
int *AllHeights[] = {Heights0, Heights1, Heights2, Heights3, Heights4, Heights5};
int *AllBlooms[] = {Blooms0, Blooms1, Blooms2, Blooms3, Blooms4, Blooms5};
int *AllWilts[] = {Wilts0, Wilts1, Wilts2, Wilts3, Wilts4, Wilts5};
int AllFlowerCounts[] = {5, 5, 5, 5, 6, 4};
printf("\n");
for(int Idx = 0;
Idx < 6;
++Idx)
{
int *Heights = AllHeights[Idx];
int *Blooms = AllBlooms[Idx];
int *Wilts = AllWilts[Idx];
int FlowerCount = AllFlowerCounts[Idx];
printf("Test %d\n", Idx);
Order(Heights, Blooms, Wilts, FlowerCount);
printf("{ ");
for(int Idx = 0;
Idx < FlowerCount;
++Idx)
{
printf("%d", Heights[Idx]);
if(Idx != (FlowerCount - 1))
{
printf(", ");
}
}
printf(" }\n\n");
}
}
编辑：这个解决方案太糟糕了，我想出了一个更好的解决方案，实际上是 DP；如下：对于每朵花，循环遍历所有其他花，检查它阻塞了哪些花；对于它挡住的那些花，检查它挡住的所有花，依此类推，直到你找到一朵不会挡住其他花的花。把那朵花放在一个新的阵列中。回溯并将每朵花放在新数组的下一个插槽中。如果对每一朵花都完成了，你会得到一个满是花的阵列，它们不会阻挡任何其他的花。然后，您将每朵花尽可能地放在前面。这个解决方案的 DP 部分是有时你会遇到之前已经被另一朵花挡住并且已经放入新数组的同一朵花，所以我们跳过那朵花而不是追赶它挡住的花。
