ZOJ - 3777 Problem Arrangement

题目描述：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3777

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The 11th Zhejiang Provincial Collegiate Programming Contest is coming! As a problem setter, Edward is going to arrange the order of the problems. As we know, the arrangement will have a great effect on the result of the contest. For example, it will take more time to finish the first problem if the easiest problem hides in the middle of the problem list.

There are N problems in the contest. Certainly, it's not interesting if the problems are sorted in the order of increasing difficulty. Edward decides to arrange the problems in a different way. After a careful study, he found out that the i-th problem placed in the j-th position will add Pij points of "interesting value" to the contest.

Edward wrote a program which can generate a random permutation of the problems. If the total interesting value of a permutation is larger than or equal to M points, the permutation is acceptable. Edward wants to know the expected times of generation needed to obtain the first acceptable permutation.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains two integers N (1 <= N <= 12) and M (1 <= M <= 500).

The next N lines, each line contains N integers. The j-th integer in the i-th line is Pij (0 <= Pij <= 100).

Output

For each test case, output the expected times in the form of irreducible fraction. An irreducible fraction is a fraction in which the numerator and denominator are positive integers and have no other common divisors than 1. If it is impossible to get an acceptable permutation, output "No solution" instead.

Sample Input

2
3 10
2 4 1
3 2 2
4 5 3
2 6
1 3
2 4

Sample Output

3/1
No solution

题意：

n个题目，每个题目在不同的位置具有不同的有趣度，求产生一个满足要求的题目排列的期望次数。

分析：

最大排列数有12！个，可以用0 1 来表示该位置是否为空

用二进制来模拟第几列已经选了，当且仅当相应的二进制位上为1

代码如下：

#include<bits/stdc++.h>
using namespace std;

int n,m;
int p[15][15],f[15];
int dp[1<<12][1300];
int gcd(int m,int n)
{
    return n ? gcd(n,m%n) : m;
}
int main()
{
    f[0] = 1;
    for(int i = 1;i <= 12;i++)
        f[i] = f[i - 1] * i;
    int t;
    scanf("%d",&t);
    memset(p,0,sizeof(p));
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= n;j++)
                scanf("%d",&p[i][j]);
        memset(dp,0,sizeof(dp));
        dp[0][0] = 1;
        for(int sta = 1;sta < (1 << n);sta++) //遍历所有状态(1 << n == pow(2,n))
        {
            int cnt = 0;
            for(int i = 1;i <= n;i++)
                if(sta & (1 << (i - 1)))
                {
                    //printf("%d---\n",sta & (1 << (i - 1)));
                    cnt++;
                }
            //printf("sta:%d cnt:%d \n",sta,cnt);
            for(int i = 1;i <= n;i++)
            {
                if( (sta & (1<<(i-1)) ) == 0 )
                    continue;
                for(int k = 0;k <= m;k++)
                {
                    if(k + p[cnt][i] >= m)
                        dp[sta][m] += dp[sta^(1<<(i-1))][k];
                    else
                        dp[sta][k + p[cnt][i] ] += dp[sta^(1<<(i-1))][k];
                }
            }
        }
        if(dp[(1 << n) - 1][m] == 0)
        {
            printf("No solution\n");
            continue;
        }
        int down = dp[(1 << n)-1][m];
        int up = f[n];
        int g = gcd(up,down);
        printf("%d/%d\n",up/g,down/g);
    }
    return 0;
}