对的最大长度链
有一对成对的链。在每对中，有两个整数，第一个整数始终较小，而第二个整数较大，同样的规则也可应用于链构建。仅当q<x时，才可以在对(p,q)之后添加对(x,y)。
为了解决这个问题，首先，我们必须按照第一个元素的升序对给定的对进行排序。之后，我们将比较一对的第二个元素和下一对的第一个元素。
输入输出
Input:
A chain of number pairs. {(5, 24), (15, 25), (27, 40), (50, 60)}
Output:
Largest length of the chain as given criteria. Here the length is 3.算法
maxChainLength(arr, n)
链中的每个元素将包含两个元素a和b
输入- 对数组，数组中的项数。
输出-最大长度。
Begin
define maxChainLen array of size n, and fill with 1
max := 0
for i := 1 to n, do
for j := 0 to i-1, do
if arr[i].a > arr[j].b and maxChainLen[i] < maxChainLen[j] + 1
maxChainLen[i] := maxChainLen[j] + 1
done
done
max := maximum length in maxChainLen array
return max
End示例
#include<iostream>
#include<algorithm>
using namespace std;
struct numPair { //define pair as structure
int a;
int b;
};
int maxChainLength(numPair arr[], int n) {
int max = 0;
int *maxChainLen = new int[n]; //create array of size n
for (int i = 0; i < n; i++ ) //Initialize Max Chain length values for all indexes
maxChainLen[i] = 1;
for (int i = 1; i < n; i++ )
for (int j = 0; j < i; j++ )
if ( arr[i].a > arr[j].b && maxChainLen[i] < maxChainLen[j] + 1)
maxChainLen[i] = maxChainLen[j] + 1;
//maxChainLen[i]现在拥有以对i结尾的最大链长
for (int i = 0; i < n; i++ )
if ( max < maxChainLen[i] )
max = maxChainLen[i]; //find maximum among all chain length values
delete[] maxChainLen; //deallocate memory
return max;
}
int main() {
struct numPair arr[] = {{5, 24},{15, 25},{27, 40},{50, 60}};
int n = 4;
cout << "Length of maximum size chain is " << maxChainLength(arr, n);
}输出结果
Length of maximum size chain is 3