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题目链接 http://poj.org/problem?id=1804
题目大意:
让一串无序数,在只能相邻数字交换的前提下,最短的次数变成有序,求该最短次数。
该最短次数=该序列的逆序数
解法1:直接双重循环求解,n*n复杂度
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
const int N = 1001;
int cyctime,len,len1,sum=0;
int arr[N];
int i=0,j=0,k=0,temp;
memset(arr,0,sizeof(int)*N);
cin >> cyctime;
for(i = 0; i < cyctime; ++i)
{
//cin.clear();
cin >> len;
len1=len;
j=0;
while(len1--) //先输入数组
{
cin >> temp;
arr[j++] = temp;
}
for(j = 0; j < len; ++j) //从前往后依次比较
{
for(k = j+1; k < len; ++k)
{
if(arr[j]>arr[k])
{
sum++;
}
}
}
cout << "Scenario #" << i+1 << ":" << endl;
cout << sum << endl << endl;
sum = 0;
}
return 0;
}
解法2:采用归并排序求解,复杂度nlgn
#include<iostream>
#include<cstring>
using namespace std;
int sum=0;
void merge(int *arr,size_t left,size_t mid,size_t right)
{
int len = right - left + 1;
int *temp = new int [len]; //数组较长时请用new,不然栈空间容易溢出
size_t index = 0;
size_t i = left, j = mid + 1;
while(i <= mid && j <= right)
{
if(arr[i]<=arr[j])
{
temp[index++] = arr[i++];
}
else
{
temp[index++] = arr[j++];
sum += mid - i + 1; //左边数比右边大,那么左边剩余的也比其大!!!!!!!!!!
}
//对两边的数组从小到大放入临时空间
}
while(i <= mid) //比较完后,左半边有没放进去的,直接写入
{
temp[index++]= arr[i++];
}
while(j <= right) //比较完后,右半边有没有放进去的,直接写入
{
temp[index++]= arr[j++];
}
for(int k = 0;k< len;++k)
{
arr[left++ ]= temp[k]; //把有序的临时数组写入原来数组的起始位置
}
delete [] temp; //释放空间
temp = NULL; //指针置空
}
void divide(int *arr,size_t left,size_t right)
{
if(left == right)
{
return;
}
size_t mid = (left+right)/2; //找出区间中部的数,将数组分段
divide(arr,left,mid); //递归调用,对左边继续分段;
divide(arr,mid+1,right); //递归调用,对右边继续分段;
merge(arr,left,mid,right); //对左右两半进行排序合并成一小段有序的数组
}
void mergesort(size_t dsize, int *arr)
{
if(dsize <= 1) //预防特殊情况下后面代码失效
{
return;
}
size_t left = 0, right = dsize-1;
divide(arr,left,right);
}
int main()
{
const int N = 1001;
int cyctime,len,len1;
int arr[N];
int i=0,j=0,temp;
memset(arr,0,sizeof(int)*N);
cin >> cyctime;
for(i = 0; i < cyctime; ++i)
{
//cin.clear();
cin >> len;
len1=len;
j=0;
while(len1--) //先输入数组
{
cin >> temp;
arr[j++] = temp;
}
mergesort(len,arr);
cout << "Scenario #" << i+1 << ":" << endl;
cout << sum << endl << endl;
sum = 0;
}
return 0;
}
由上可看出归并排序求解时间效率更高。