题目地址:POJ 1731 这题能够直接用STL函数做,非常轻松..next_permutation函数非常给力.. 代码例如以下: #include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <queue> #include <cmath> #include <st…

http://poj.org/problem?id=1731 Orders Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9702   Accepted: 5925 Description The stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having label…

题目链接:http://poj.org/problem?id=1731 思路分析:含有重复元素的全排列问题:元素个数为200个,采用暴力枚举法. 代码如下: #include <iostream> #include <algorithm> using namespace std; + ; void PrintPermu( int n, char P[], char A[], int cur ) { int i, j; if ( cur == n ) { ; i < n; ++…

The stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having labels starting with the same letter are stored in the same warehouse (i.e. in the same building) labelled with this letter. During the da…

Orders Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9940   Accepted: 6048 Description The stores manager has sorted all kinds of goods in an alphabetical order of their labels. All the kinds having labels starting with the same letter…

Expedition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8053   Accepted: 2359 Description A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to…

题意: 给你n*m的矩阵,然后每行取一个元素,组成一个包含n个元素的序列,一共有n^m种序列, 让你求出序列和最小的前n个序列的序列和. 解题思路: 1.将第一序列读入seq1向量中,并按升序排序. 2.将数据读入seq2向量中,并按升序排序. 将seq2[0] +seq1[i] ( 0<=i<=n-1)读入seqn向量中 用make_heap对seqn建堆. 然后seq2[1] + seq1[i] (0<=i<=n-1),如果seq2[1] +seq1[i]比堆seqn的顶点大,…

排列 Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15173   Accepted: 6148 Description 题目描述: 大家知道,给出正整数n,则1到n这n个数可以构成n!种排列,把这些排列按照从小到大的顺序(字典顺序)列出,如n=3时,列出1 2 3,1 3 2,2 1 3,2 3 1,3 1 2,3 2 1六个排列. 任务描述: 给出某个排列,求出这个排列的下k个排列,如果遇到最后一个排列,则下1排列为…

#include<iostream> #include<string> #include<algorithm> using namespace std; int main() { int NUM; string list; cin>>list,NUM = list.length(),sort(&list[],&list[]+NUM),cout<<list<<endl; ],&list[]+NUM))cout&l…

先把不是标准格式的字符串变成标准格式再输出出现两次以上的标准串和出现的次数不然输出 "No duplicates." #include <iostream> #include <cstring> #include <string> #include <map> #include <vector> #include <algorithm> using namespace std; int n; map<char…