1011 - Marriage Ceremonies PDF (English) Statistics ForumTime Limit: 2 second(s) Memory Limit: 32 MBYou 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 give priorities to family background, some give priorities to education, etc.
Now your company is in a danger and you want to save your company from this financial crisis by arranging as much marriages as possible. So, you collect N bio-data of men and N bio-data of women. After analyzing quite a lot you calculated the priority index of each pair of men and women.
Finally you want to arrange N marriage ceremonies, such that the total priority index is maximized. Remember that each man should be paired with a woman and only monogamous families should be formed.
InputInput starts with an integer T (≤ 100), denoting the number of test cases.
Each case contains an integer N (1 ≤ n ≤ 16), denoting the number of men or women. Each of the next N lines will contain N integers each. The jth integer in the ith line denotes the priority index between the ith man and jth woman. All the integers will be positive and not greater than 10000.
OutputFor each case, print the case number and the maximum possible priority index after all the marriages have been arranged.
Sample InputOutput for Sample Input221 52 131 2 36 5 48 1 2Case 1: 7Case 2: 16
解题思路:用dp[i][j] 来表示，其中i是表示计算到第几层，j的二进制中的1表示已经取了那些位置的数字
状态转移方程dp[i][j|(1<<k)] = max(dp[i][j],dp[i-1][j],map[i-1][k]);
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; int map[20][20]; int dp[20][65635]; int T,n,len; int main(){ scanf("%d",&T); for(int t=1;t<=T;t++){ memset(dp,0,sizeof(dp)); scanf("%d",&n); len = (int)pow(2,n); for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%d",&map[i][j]); for(int i=1;i<=n;i++){ for(int j=0;j<n;j++){ for(int k=0;k<len;k++){ if(((1<<j)&k)) continue; dp[i][(1<<j)|k] = max(dp[i][(1<<j)|k],dp[i-1][k]+map[i-1][j]); } } } printf("Case %d: %d\n",t,dp[n][len-1]); } return 0; }
转载于:https://www.cnblogs.com/yuanshixingdan/p/5560418.html
