Made In Heaven 题目描述 One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are N spots in the jail and M roads connecting some of the spots. JOJO f…

思路:K短路裸题 代码: #include<queue> #include<cstring> #include<set> #include<map> #include<stack> #include<string> #include<cmath> #include<vector> #include<cstdio> #include<iostream> #include<algori…

One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are NN spots in the jail and MM roads connecting some of the spots. JOJO finds that Pucci kn…

Lattice's basics in digital electronics 题目链接 题目描述 LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10−9 second. In the next 10−9 second, he built a data decoding device that decodes…

Supreme Number 题目链接 题目描述 A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers. Now lets define a number N as the supreme number if and only if each number made up of an non-em…

Made In Heaven One day in the jail, F·F invites Jolyne Kujo (JOJO in brief) to play tennis with her. However, Pucci the father somehow knows it and wants to stop her. There are N spots in the jail and MM roads connecting some of the spots. JOJO finds…

面试题 15:链表中倒数第 k 个结点 题目:输入一个链表,输出该链表中倒数第 k 个结点.为了符合大多数人的习惯, 本题从 1 开始计数,即链表的尾结点是倒数第一个结点.例如一个有 6 个结点的 链表,从头结点依次是 1,2,3,4,5,6.倒数第三个结点就是值为 4 的结点. 最初思路:先遍历链表得到链表总长度,然后length-k+1位置的结点就是所求结点.但是需要遍历两遍 改进:用两个指针,两个指针之间距离相差k-1;当第一个指针都到链表尾结点,那么第二个指针就在倒数第k 注意程序的鲁棒…

///////////////////////////////////////////////////////////////////////////////////// // 5. 题目15 链表中倒数第K个节点 //时间复杂度:O(n),空间复杂度O(n) ListNode<int>* KthNodeFromEnd(ListNode<int>* pNode, int k) { ) { return NULL; } ListNode<int>* p = pNode;…

https://nanti.jisuanke.com/t/31445 题意 能否在t时间内把第k短路走完. 分析 A*算法板子. #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <cti…

求第k短路模板 先逆向求每个点到终点的距离,再用dij算法,不会超时(虽然还没搞明白为啥... #include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<vector> #include<string.h> #include<cstring> #include<algorithm> #include<set&g…