首页 > ACM题库 > HDU-杭电 > hdu 2659 Cheat待解决[解题报告]C++
2014
02-12

hdu 2659 Cheat待解决[解题报告]C++

Cheat

问题描述 :

In the summer of last year, Lemon, Yifenfei and CTW live in the same apartment. They usually played some game after they came back from lab. One day, Yifenfei came up with an idea about a new game. That game’s rules are as follows:
Yifenfei provides a positive integer n, Lemon and CTW should give 3 positive integers x, y and z in a limited time. The winner is who make the |n – x * y * z| as small as possible. If the value of |n – x * y * z| is tied, the rusult is determined by smaller x, then smaller y, then smaller z, until the winner appear. To make this game more diffcult, Yifenfei will also give a list of positive integers, x, y and z cannot be included in this list.
Because Lemon is so smart, CTW never win even one game. Can you provides CTW a program to help him to get the optimum answer?

输入:

Input includes several cases. The first line of a case contains two integers t, n. Then follows t positive integers in the seconde line.Input ends with a case t = 0 and n = 0. 0 <= t <= 50, 0 < n <= 1000.

输出:

Input includes several cases. The first line of a case contains two integers t, n. Then follows t positive integers in the seconde line.Input ends with a case t = 0 and n = 0. 0 <= t <= 50, 0 < n <= 1000.

样例输入:

1 1
1
2 10
1 2
0 0

样例输出:

2 2 2
3 3 3


  1. 在方法1里面:

    //遍历所有的边,计算入度
    for(int i=0; i<V; i++)
    {
    degree = 0;
    for (j = adj .begin(); j != adj .end(); ++j)
    {
    degree[*j]++;
    }
    }

    为什么每遍历一条链表,要首先将每个链表头的顶点的入度置为0呢?
    比如顶点5,若在顶点1、2、3、4的链表中出现过顶点5,那么要增加顶点5的入度,但是在遍历顶点5的链表时,又将顶点5的入度置为0了,那之前的从顶点1234到顶点5的边不是都没了吗?