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You have been given the job of forming the quiz teams for the next MCA CPCI Quiz Championship There are*N students interested to participate and you have to form N teams each team consisting of two members Since the members have to practice together all the students want their members house as near as possible Let x be the distance between the houses of group x be the distance between the houses of group and so on You have to make sure the summation (x + x + x + … + xn) is minimized Input There will be many cases in the input file Each case starts with an integer N (N ≤ ) The next *Nlines will given the information of the students Each line starts with the students name followed by thex coordinate and then the y coordinate Both x y are integers in the range to Students name will consist of lowercase letters only and the length will be at most Input is terminated by a case where N is equal to Output For each case output the case number followed by the summation of the distances rounded to decimal places Follow the sample for exact format 题意给定n有*n个人分成两人一组的n组每个人在一个坐标上要求出所有组的两人距离的和最小 思路明显的集合最优配对问题i最为前i个人s作为前i个人能组成的所有集合j为前i个人中的一个人 这样状态转移方程为dp[i][s] =min(dp[i][s] dis(i j) + dp[i ][s {i} {j}]集合用二进制来表示所以方程变为 dp[i][s] =min(dp[i][s] dis(i j) + dp[i ][s ^ <<i ^ <<j]但是这样做时间不是很理想然后看了个详细解法 里面讲得挺详细的可以转化为维 代码 #include <stdioh>#include <stringh>#include <limitsh>#include <mathh> int n i j s;double dp[<<];struct Student{ char name[]; int x y;} stu[]; double min(double a double b) { return a < b ? a : b;} double dis(Student a Student b) { return sqrt((double)((ax bx) * (ax bx) + (ay by) * (ay by)));} double dpp(int p) { if (dp[p] != ) return dp[p]; dp[p] = INT_MAX; int i j; for (i = n ; i >= ; i ) if (p & ( << i)) break; for (j = i ; j >= ; j ) if (p & ( << j)) dp[p] = min(dp[p] dis(stu[i] stu[j]) + dpp(p ^ (<<i) ^ (<<j))); return dp[p];}int main(){ int t = ; while (~scanf(%d &n) && n) { n *= ; s = <<n; for (i = ; i < s; i ++) dp[i] = ; dp[] = ; for (i = ; i < n; i ++) scanf(%s%d%d stu[i]name &stu[i]x &stu[i]y); printf(Case %d: %lfn t ++ dpp(s )); } return ;} |
