[英]Calculating Execution time and big O time of my c++ program
我的代碼是:
#include <iostream>
#include <utility>
#include <algorithm>
//#include <iomanip>
#include <cstdio>
//using namespace std;
inline int overlap(std::pair<int,int> classes[],int size)
{
std::sort(classes,classes+size);
int count=0,count1=0,count2=0;
int tempi,tempk=1;
for(unsigned int i=0;i<(size-1);++i)
{
tempi = classes[i].second;
for(register unsigned int j=i+1;j<size;++j)
{
if(!(classes[i].first<classes[j].second && classes[i].second>classes[j].first))
{ if(count1 ==1)
{
count2++;
}
if(classes[i].second == tempi)
{
tempk =j;
count1 = 1;
}
////cout<<"\n"<<"Non-Overlapping Class:\t";
////cout<<classes[i].first<<"\t"<<classes[i].second<<"\t"<<classes[j].first<<"\t"<<classes[j].second<<"\n";
classes[i].second = classes[j].second;
count++;
if(count1==1 && j ==(size-1))
{
j= tempk;
classes[i].second = tempi;
count1= 0;
if(count2 !=0)
{
count = (count + ((count2)-1));
}
count2 =0;
}
}
else
{
if(j ==(size-1))
{
if(count>0)
{
j= tempk;
classes[i].second = tempi;
count1= 0;
if(count2 !=0)
{
count = (count + ((count2)-1));
}
count2 =0;
}
}
}
}
}
count = count + size;
return count;
}
inline int fastRead_int(int &x) {
register int c = getchar_unlocked();
x = 0;
int neg = 0;
for(; ((c<48 || c>57) && c != '-'); c = getchar_unlocked());
if(c=='-') {
neg = 1;
c = getchar_unlocked();
}
for(; c>47 && c<58 ; c = getchar_unlocked()) {
x = (x<<1) + (x<<3) + c - 48;
}
if(neg)
x = -x;
return x;
}
int main()
{
int N;
////cout<<"Please Enter Number Of Classes:";
clock_t begin,end;
float time_interval;
begin = clock();
while(fastRead_int(N))
{
switch(N)
{
case -1 : end = clock();
time_interval = float(end - begin)/CLOCKS_PER_SEC;
printf("Execution Time = %f",time_interval);
return 0;
default :
unsigned int subsets;
unsigned int No = N;
std::pair<int,int> classes[N];
while(No--)
{
////cout<<"Please Enter Class"<<(i+1)<<"Start Time and End Time:";
int S, E;
fastRead_int(S);
fastRead_int(E);
classes[N-(No+1)] = std::make_pair(S,E);
}
subsets = overlap(classes,N);
////cout<<"\n"<<"Total Number Of Non-Overlapping Classes is:";
printf("%08d",subsets);
printf("\n");
break;
}
}
}
和我的程序的輸入和輸出:
Input:
5
1 3
3 5
5 7
2 4
4 6
3
500000000 1000000000
1 5
1 5
1
999999999 1000000000
-1
Output:
Success time: 0 memory: 3148 signal:0
00000012
00000005
00000001
Execution Time = 0.000036
我試圖通過在main的開始和main的末尾都有時鍾來計算並找出時間,但是它只說了0.000036秒。但是當我試圖在Online Judge(SPOJ)中發布相同的代碼時,我的程序得到了'Time超出限制”錯誤。 SPOJ中上述程序的時間限制為2.365秒。有人可以幫我解決這個問題嗎?
我認為您的問題與overlap
功能有關。
里面有
for
循環:
0..Size
i..Size
第二個循環的內部稱為Size(Size + 1)/ 2(N個第一個整數的倒數和),沒有中斷。
因此,您的算法是O(n²),其中n是大小。
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