help.cpp
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/*************************************************************
/* name: help.cpp
/* Description: 辅助函数集合
/* Purpose：满足以下条件，合并两个框：
/* 1 大框中包含小框 2 框交叉 3 框相邻
/* Update Time: by zl 20160307
**************************************************************/
#include "help.h"
#include <math.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
//排序
bool SortByPoint(const CForeground &C1,const CForeground &C2)
{
bool sign = false;
if(C1.m_rtConnect.left < C2.m_rtConnect.left)
return true;
else if (C1.m_rtConnect.left == C2.m_rtConnect.left)
{
if(C1.m_rtConnect.top <= C2.m_rtConnect.top)
return true;
else
return false;
}
else
return false;
}
//判断x是否在边界内
bool InRange(int x, int boundary1, int boundary2)
{
int max_b = boundary1 >= boundary2 ? boundary1 : boundary2;
int min_b = boundary1 < boundary2 ? boundary1 : boundary2;
if (x <= max_b && x >= min_b)
return true;
else
return false;
}
//相邻或者包含
bool NeighbourorSub(CForeground c1, CForeground c2)
{
//c2中心点在c1内部 需要合并 返回true
/*if (c2.m_ptCenter.x < c1.m_rtConnect.right && InRange(c2.m_ptCenter.y, c1.m_rtConnect.top, c1.m_rtConnect.bottom))
return true;
else */
//输入条件为c2一定在c1的右侧 因此若交叉或者包含 c2的左下点 或者 右上点 肯定至少有一个在c1的内部
if (c2.m_rtConnect.left <= c1.m_rtConnect.right)
{
//包含或者交叉
if (InRange(c2.m_rtConnect.bottom, c1.m_rtConnect.top, c1.m_rtConnect.bottom) || InRange(c2.m_rtConnect.top, c1.m_rtConnect.top, c1.m_rtConnect.bottom))
return true;
//上方邻近 或者 下方邻近
else if (abs(c1.m_rtConnect.top - c2.m_rtConnect.bottom) < MergeMinDistance || abs(c1.m_rtConnect.bottom - c2.m_rtConnect.top) < MergeMinDistance)
return true;
}
else if (InRange(c2.m_rtConnect.bottom, c1.m_rtConnect.top, c1.m_rtConnect.bottom) || InRange(c2.m_rtConnect.top, c1.m_rtConnect.top, c1.m_rtConnect.bottom))
{
if((abs(c1.m_rtConnect.right - c2.m_rtConnect.left) < MergeMinDistance)) //c2 在 c1 右侧 邻近
return true;
}
return false;
}
//合并
void MergeNeighborBox(vector<CForeground> &ForegroundArray)
{
//printf("\n ====== MergeNeighborBox =========\n");
vector<CForeground>::iterator iter1 = ForegroundArray.begin();
for(; iter1 != ForegroundArray.end() - 1 && iter1 != ForegroundArray.end(); iter1++)
{
//printf("%d %d %d %d\n", iter1->m_rtConnect.left, iter1->m_rtConnect.top, iter1->m_rtConnect.right, iter1->m_rtConnect.bottom);
vector<CForeground>::iterator iter2 = iter1 + 1;
for (; iter2 != ForegroundArray.end(); )
{
//printf(" %d %d %d %d\n", iter2->m_rtConnect.left, iter2->m_rtConnect.top, iter2->m_rtConnect.right, iter2->m_rtConnect.bottom);
//if ((x2 <= iter1->m_rtConnect.right && y2 <= iter1->m_rtConnect.bottom && y2 >= iter1->m_rtConnect.top) || //大框含小框状况
// (iter2->m_rtConnect.left <= iter1->m_rtConnect.right && iter2->m_rtConnect.bottom <= iter1->m_rtConnect.bottom && iter2->m_rtConnect.bottom >= iter1->m_rtConnect.top) || //框交叉 左上角在第一个框内
// (iter2->m_rtConnect.left <= iter1->m_rtConnect.right && iter2->m_rtConnect.top <= iter1->m_rtConnect.bottom && iter2->m_rtConnect.top >= iter1->m_rtConnect.top) ||//框交叉 左下角在第一个框内
// iter2->m_rtConnect.left <= iter1->m_ptCenter.x && abs(iter2->m_rtConnect.bottom - iter1->m_rtConnect.top) < 10 ||//距离较近的
// iter2->m_rtConnect.left <= iter1->m_ptCenter.x && abs(iter2->m_rtConnect.top - iter1->m_rtConnect.bottom) < 10 ||
// distance < 20 //距离较近的
// )
if (NeighbourorSub(*iter1, *iter2))
{
//合并
iter1->m_rtConnect.left = min(iter1->m_rtConnect.left, iter2->m_rtConnect.left);
iter1->m_rtConnect.bottom = max(iter1->m_rtConnect.bottom, iter2->m_rtConnect.bottom);
iter1->m_rtConnect.right = max(iter1->m_rtConnect.right, iter2->m_rtConnect.right);
iter1->m_rtConnect.top = min(iter1->m_rtConnect.top, iter2->m_rtConnect.top);
iter1->m_ptCenter.x = (iter1->m_rtConnect.right + iter1->m_rtConnect.left) / 2;
iter1->m_ptCenter.y = (iter1->m_rtConnect.bottom + iter1->m_rtConnect.top) / 2;
//删除一个
iter2 = ForegroundArray.erase(iter2);
iter1 = iter2;
iter1--;
}
else
iter2++;
//else
// break; //由于ForegroundArray按x坐标排列 所以认为如果当前可以合并，则有可能下一个也需要合并；
//但若当前不符合合并 则下一个也不会符合（为了减少计算次数）
}
}
}
int ifOverlapBox(CForeground iter1, CForeground iter2)
{
if (iter1.m_rtConnect.left <= iter2.m_rtConnect.left &&iter1.m_rtConnect.top <= iter2.m_rtConnect.top
&& iter1.m_rtConnect.right >= iter2.m_rtConnect.right && iter1.m_rtConnect.bottom >= iter2.m_rtConnect.bottom)
return 2;
else if (iter1.m_rtConnect.left >= iter2.m_rtConnect.left &&iter1.m_rtConnect.top >= iter2.m_rtConnect.top
&& iter1.m_rtConnect.right <= iter2.m_rtConnect.right && iter1.m_rtConnect.bottom <= iter2.m_rtConnect.bottom)
return 1;
else return 0;
}
//合并之后 去除完全重叠的
void DislodgeOverlapBox(vector<CForeground> &ForegroundArray)
{
vector<CForeground>::iterator iter1 = ForegroundArray.begin();
for (; iter1 != ForegroundArray.end() - 1 && iter1 != ForegroundArray.end(); iter1++)
{
//printf("%d %d %d %d\n", iter1->m_rtConnect.left, iter1->m_rtConnect.top, iter1->m_rtConnect.right, iter1->m_rtConnect.bottom);
vector<CForeground>::iterator iter2 = iter1 + 1;
for (; iter2 != ForegroundArray.end(); )
{
//printf(" %d %d %d %d\n", iter2->m_rtConnect.left, iter2->m_rtConnect.top, iter2->m_rtConnect.right, iter2->m_rtConnect.bottom);
int overlap_box = ifOverlapBox(*iter1, *iter2);
if (!overlap_box)
{
iter2++;
continue;
}
else if (overlap_box == 1) //iter1 包围 iter2，删除iter2
{
iter2 = ForegroundArray.erase(iter2);
}
else ////iter2 包围 iter1，删除iter1
{
iter1 = ForegroundArray.erase(iter1);
break;
}
}
}
}
int checkTime()
{
struct tm* info;
int nYear, nMonth, nDay;
time_t raw;
time(&raw);
info = localtime(&raw);
nYear = info->tm_year + 1900;
nMonth = info->tm_mon + 1;
nDay = info->tm_mday;
if (nYear != 2020)//2017年1月1日版本过期
{
printf("版本过期！\n");
return -1;
}
else
return 1;
}