奋斗青年的专栏

原题地址 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=94#include #include #include using namespace std;class matrixC{    public:        char name;        int row,col;    ma

题目描述你有一些小球，从左到右依次编号为1,2,3,...,n. 你可以执行两种指令（1或者2）。其中， 1 X Y表示把小球X移动到小球Y的左边， 2 X Y表示把小球X移动到小球Y右边。 指令保证合法，即X不等于Y。 例如，初始状态1,2,3,4,5,6的小球执行1 1 4后，小球1被移动到小球4的左边，即2,3,1,4,5,6。如果再执行2 3 5，结点3将会移到5的右边，即2,1,

#include#include using namespace std;struct NodeT{ int n; NodeT *next,*front;};NodeT* kill(NodeT *p,int Kn){ //从给你的指针位置开始数1 ，一直数到 Kn //把那个人杀掉，并且返回被杀掉的后一个人的指针。 NodeT *pre=p;

题目描述：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并输出它的后序遍历序列。输入：输入可能包含多个测试样例，对于每个测试案例，输入的第一行为一个整数n(1输入的第二行包括n个整数(其

The Blocks ProblemTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4400 Accepted: 1822DescriptionMany areas of Computer Science use simple, abstract doma

The Dole Queue In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decided on the following strategy. Every day all dole applicants

虽然划分在数据结构里，但感觉什么都没用到。简单的字符串处理罢了。

Parentheses Balance You are given a string consisting of parentheses () and []. A string of this type is said to be correct:(a)if it is the empty string(b)if A and B are correct, AB is c

用#include #include #include #include #include #include #include using namespace std;int height,width,cnt;int map[240][240];int ans,times=0;int step_x[]={1,-1,0,0};int step_y[]={0,0,1,-1};

#include #include #include using namespace std;struct num_strT{ //0代表数字 1 代表字符 bool flag; int num; char str;};class stack1{ public: char wait[1005]; int count;

广度优先搜索，超级好玩，

用一个Dps，把它输出来。A quadtree is a representation format used to encode images. The fundamental idea behind the quadtree is that any image can be split into four quadrants. Each quadrant may again be spli

#include #include #include using namespace std;int main(){    stack myStack;    char s[1002],*p,temp;    int j,n;    while(cin>>s){    n=strlen(s);    for(j=0;j    {        i

Generalized MatrioshkasTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1816 Accepted: 528DescriptionVladimir worked for years making matrioshkas, thos

一道树的题目#include #include#include#include#include#include#include#include#include#include#includeusing namespace std;struct node{ char data; node*left; node*right;};node*

Tree SummingTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7866 Accepted: 1815DescriptionLISP was one of the earliest high-level programming languages

Team Queue Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday lif

Problem Description某部队进行新兵队列训练，将新兵从一开始按顺序依次编号，并排成一行横队，训练的规则如下：从头开始一至二报数，凡报到二的出列，剩下的向小序号方向靠拢，再从头开始进行一至三报数，凡报到三的出列，剩下的向小序号方向靠拢，继续从头开始进行一至二报数。。。，以后从头开始轮流进行一至二报数、一至三报数直到剩下的人数不超过三人为止。  

胜利大逃亡Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21542    Accepted Submission(s): 8479Problem DescriptionIgnatius被魔王抓走了,

重建一棵树  Tree You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any le

#include #include #include #include #include using namespace std;vectorv;int total;vector::iterator itr;char alph[10]={'#','a','b','c','d','e','f','g','h'};void dps(int n){ // 这里添加进来的是 元