Marriage Ceremonies LightOJ - 1011 常规状压dp.popcount(S)表示S集合中元素数量.ans[S]表示S中的女性与前popcount(S)个男性结婚的最大收益. 那么,$ans[S]=max\{ans[S-p]+a[popcount(S)][p]\}$ 老代码 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ][],a[][]…

http://www.lightoj.com/volume_showproblem.php?problem=1011 You work in a company which organizes marriages. Marriages are not that easy to be made, so, the job is quite hard for you. The job gets more difficult when people come here and give their bi…

You work in a company which organizes marriages. Marriages are not that easy to be made, so, the job is quite hard for you. The job gets more difficult when people come here and give their bio-data with their preference about opposite gender. Some gi…

 Marriage Ceremonies Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1011 Description You work in a company which organizes marriages. Marriages are not that easy to be made, so, the job is qu…

1011 - Marriage Ceremonies   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB You work in a company which organizes marriages. Marriages are not that easy to be made, so, the job is quite hard for you. The job gets more diff…

题目大意: 有N个男人,和N个女人要互相匹配,每个男人和每个女人有个匹配值. 并且匹配只能是1对1的. 问所有人都匹配完成,最大的匹配值是多少?   状压DP,暴力枚举就OK了, 这个题目略坑,因为他卡常数,可以有一些简单的优化,能优化1500ms............. =================================================================   #include<cstdio> #include<cstring> #i…

题目大意: 给出n*n的矩阵Map,Map[i][j]代表第i个男人和第j个女人之间的满意度,求男女一一配对后,最大的满意度之和. 题目思路:状态压缩 题目可看做每行取一点,所有点不同列的情况下,各个点的最大和为多少. dp[i][j],代表第i行,状态为j的情况下的最优解,其中j的含义为:j所代表的二进制数中:若第i位为1则取第i列的值,为0则不取. 这样j从1到(1<<n)-1,便包含了矩阵中所有列的取法 为了节约时间,我们需要作出一点优化:我们知道i*i的矩阵中不可能存在比i大的列,也就…

题目链接 大概题意是有n个男的n个女的(原谅我这么说,我是粗人),给你一个n*n的矩阵,第i行第j列表示第i个女(男)对第j个男(女)的好感度,然后要安排n对相亲,保证都是正常的(无搞基百合之类的),然后求怎么安排能使好感度和最大,求出最大值. 开始试了纯暴力的方法,时间复杂度是n!果断超时 #include <iostream> #include <string.h> #include <algorithm> using namespace std; int vis[…

由于暂时不会KM算法,只能用最大费用流来做了. 题目链接:http://lightoj.com/volume_showproblem.php?problem=1011 #include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> #include <queue> #include <vector>…

思路:状态压缩dp,设dp[i][j] 表示前i行,状态为j时的最大值,状态定义为:若前i行中取了第x列那么j的二进制位中第x位为1,否则为0,最后答案就是dp[n-1][(1 << n)-1]; 装态转移方程就是 dp[i][j|(1 << k)] = max(dp[i][j|(1 << k)],dp[i-1][j] + mat[i][k])  (j & (1 << k) == 0) #include<cstdio> #include&…