题目链接:10911 - Forming Quiz Teams 题目大意:给出2 * n个选手的坐标, 要求将所有的选手分成n组, 每组两个人, 所有组的两个人之间的距离之和要最小, 输出最小值. 解题思路:网络赛的时候写过类似的题目, 只不过是选4个点做正方形,所以思路很明确,每次选取任意两个点配对,递归搜索,并记录下来.然后我不是用未运算来记录点的状态,而开了个数组,因为位运算用不熟. #include <stdio.h> #include <string.h> #includ…

4th IIUC Inter-University Programming Contest, 2005 G Forming Quiz Teams Input: standard input Output: standard output Problemsetter: Sohel Hafiz You have been given the job of forming the quiz teams for the next ‘MCA CPCI Quiz Championship’. There a…

题目大意:给出2*n个点,将这些点配成n对,使得所有点对中两点的距离之和尽量小. 用一个整数的二进制形式表示一个集合的子集,以每个集合为状态进行状态转移.具体参见<算法竞赛入门经典>9.5. #include <cstdio> #include <cmath> #include <algorithm> #include <cfloat> using namespace std; struct Point { int x, y; } point[]…

/* 提醒推荐:五星 刘汝佳<算法竞赛入门经典>,集合上的动态规划---最优配对问题 题意:空间里有n个点P0,P1,...,Pn-1,你的任务是把它们配成n/2对(n是偶数),使得每个点恰好在一个点对中.所有点对中两点的距离之和应尽量小. 状态:d(i,S)表示把前i个点中,位于集合S中的元素两两配对的最小距离和 状态转移方程为:d(i,S)=min{|PiPj|+d(i-1,S-{i}-{j}} 书上的解法有些问题,正解见方法一 方法二:状态可以进行压缩,i的值其实隐藏在S中,S中最高位为…

n个人里选k个人有C(n, k)中方法,再从里面选一人当队长,有k中方法. 所以答案就是 第一步的变形只要按照组合数公式展开把n提出来即可. #include <cstdio> typedef long long LL; ; LL pow(int p) { LL ans = , ; while(p) { ) ans = (ans * base) % M; p >>= ; base = (base * base) % M; } return ans; } int main() { /…

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E. Two Teams time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are nn students standing in a row. Two coaches are forming two teams — the first coach chooses the first team and the sec…

10192 最长公共子序列 http://uva.onlinejudge.org/index.php?option=com_onlinejudge& Itemid=8&page=show_problem&category=114&problem=1133&mosmsg= Submission+received+with+ID+13297616 */ #include <cstdio> #include <string.h> #include&…

Rainbow 6 is a very popular game in colleges. There are 2 teams, each having some members and the 2 teams play some matches against each other. The team which wins the maximum number of matches wins the game! Two of my friends Ashank and Aditya (bett…