【搜索】POJ-2718 全排列+暴力

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【搜索】POJ-2718 全排列+暴力

一.题目 Description Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some order to form a second integer. Unles…

POJ 2718【permutation】

POJ 2718 问题描述: 给一串数,求划分后一个子集以某种排列构成一个数,余下数以某种排列构成另一个数,求这两个数最小的差,注意0开头的处理. 超时问题:一开始是得到一个数列的组合之后再从中间进行切割得到两数,会超时.后来采用的方法是将前面的数在DFS中得到固定,在函数work中对后面(n-n/2)个数进行排列组合. 针对整个数列的dfs排列组合剪枝,若当前搜索的第二个数后面全部补零与第一个数所产生的差值比当前所搜索到的结果还要大,那么就直接返回.这个剪枝就是超时与几十MS的差距int nu…

POJ 2718 Smallest Difference（最小差）

Smallest Difference(最小差) Time Limit: 1000MS Memory Limit: 65536K Description - 题目描述 Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining d…

poj 2718 Smallest Difference(暴力搜索+STL+DFS)

Smallest Difference Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6493 Accepted: 1771 Description Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in…

穷竭搜索： POJ 2718 Smallest Difference

题目:http://poj.org/problem?id=2718 题意: 就是输入N组数据,一组数据为,类似 [1 4 5 6 8 9]这样在0~9之间升序输入的数据,然后从这些数据中切一刀,比如 n1:[1 4 5],n2:[6 8 9]这样,然后 abs(n1- n2),对n1 和 n2的所有可能的排列 n1: [1 4 5][1 5 4]...这样,要算出来的最小的差,显然从中间切一刀才会出现这种解. 题解: 这里可以用来练习 STL,算法不会也没有关系,可以在这题学到 bi…

[暴力搜索] POJ 3087 Shuffle'm Up

Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10003 Accepted: 4631 Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks o…

poj 2718 Smallest Difference(穷竭搜索dfs)

Description Given a number of distinct , the integer may not start with the digit . For example, , , , , and , you can write the pair of integers and . Of course, there are many ways to form such pairs of integers: and , and , etc. The absolute value…