计算完美HASH

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原创最新推荐文章于 2026-07-08 10:06:17 发布 718 阅读 AI 写同款· GEO 优化 ›

391. 完美矩形本题主要涉及到对多个矩形的几何关系的判断。可以通过计算所有矩形的面积之和，并与最终覆盖的矩形区域的面积进行比较，同时检查矩形的顶点是否符合覆盖条件来解决。表示一个坐标轴平行的矩形。这个矩形的左下顶点是。如果所有矩形一起精确覆盖了某个矩形区域，则返回。阅读详情

// 计算完美HASH

#include "stdafx.h"
#include <windows.h>
#include <algorithm>
#include <iostream>

using namespace std;

// HASH函数为 data%HashSize
// 返回完美HASH的尺寸
// 失败返回0
unsigned int GetPerfectHash(const unsigned int* pData, size_t iNum, size_t iMaxHashSize);

int _tmain(int argc, _TCHAR* argv[])
{
unsigned int data[] = {204, 83, 111, 103, 46, 115, 82, 145};
size_t nNum = sizeof(data)/sizeof(data[0]);
size_t iHashSize = GetPerfectHash(data, nNum, nNum*3);
cout << "// HashSize = " << iHashSize << endl;
if(iHashSize != 0)
{
  for(size_t i=0; i<nNum; ++i)
  {
   cout << "// hash(" << data[i] << ") = " << data[i]%iHashSize << endl;
  }
}

return 0;
}

unsigned int GetPerfectHash(const unsigned int* pData, size_t iNum, size_t iMaxSize)
{
if(pData == 0)
return 0;

if(iNum <= 1)
return iNum;

if(iMaxSize < iNum)
iMaxSize = iNum;

bool* pConflict = new bool[iMaxSize];

for(size_t nHashSize = iNum; nHashSize<=iMaxSize; ++nHashSize)
{
fill(pConflict, pConflict+nHashSize, false);

  bool bFindPerfect = true;
  for(size_t i=0; i<iNum; ++i)
  {
   unsigned int iHash = pData[i] % nHashSize;
   if(pConflict[iHash])
   {
    bFindPerfect = false;
    break;
   }
   pConflict[iHash] = true;
  }

  if(bFindPerfect)
  {
   delete[] pConflict;
   return nHashSize;
  }
}

delete[] pConflict;
return 0;
}

// 运行结果:
// HashSize = 22
// hash(204) = 6
// hash(83) = 17
// hash(111) = 1
// hash(103) = 15
// hash(46) = 2
// hash(115) = 5
// hash(82) = 16
// hash(145) = 13