ACM学习历程—ZOJ 3777 Problem Arrangement（递推 && 状压）

Description

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 addP_ij 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 P_ij (0 <= P_ij <= 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

题目大意跟八皇后很像，每行每列只取一个，然后求和，要求大于等于m的概率。

首先根据乘法原理，一共有n！种取法。也就是最多12! = 479001600这个复杂度太大。

但是这么多状态都是互异的，是不可能不计算的。

于是考虑状态能不能合并，考虑到我第一行取第一个，第二行取第三个这种情况，和第一行取第三个，第二行取第一个这种情况，都导致后面的行不能取1、3两列。

于是从第一行开始取，只考虑哪几列取过了。于是p[state][w]就表示取了state(二进制状压)的状态下，和为w的种数。

那么p[state|(1<<i)][w+a[cnt+1][i]] += p[state][w];

cnt表示当前取过几行，i表示那一列没有取过。

这样的话递推关系就能实现了。

最后要求大于等于m的减一下就出来了。

时间复杂度：O(n*m*2^n)

最大：12*500*2^12 = 24576000降了一个数量级。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <algorithm>
#define LL long long

using namespace std;

typedef pair<int, int> pii;
int n, m, a[][];
int p[(<<)+][], to, all;
bool vis[(<<)+];

void input()
{
    scanf("%d%d", &n, &m);
    for (int i = ; i <= n; ++i)
        for (int j = ; j <= n; ++j)
            scanf("%d", &a[i][j]);
    memset(p, , sizeof(p));
    memset(vis, false, sizeof(vis));
    p[][] = ;
    to = ;
    all = ;
    for (int i = ; i <= n; ++i)
    {
        to |= (<<i);
        all *= i;
    }
}

//GCD
//求最大公约数
//O(logn)
int gcd(int a, int b)
{
    if (b == )
    return a;
    else
    return gcd(b, a%b);
}

void bfs()
{
    queue<pii> q;
    q.push(pii(, ));
    vis[] = true;
    pii now;
    int k, cnt;
    while (!q.empty())
    {
        now = q.front();
        q.pop();
        k = now.first;
        cnt = now.second;
        vis[k] = false;
        for (int i = ; i <= n; ++i)
        {
            if (k&(<<i))
                continue;
            for (int v = ; v <= m; ++v)
            {
                if (p[k][v] == )
                    continue;
                p[k|(<<i)][v+a[cnt+][i]] += p[k][v];
                if (!vis[k|(<<i)] && cnt+ != n)
                {
                    q.push(pii(k|(<<i), cnt+));
                    vis[k|(<<i)] = true;
                }
            }
        }
    }
}

void work()
{
    bfs();
    int ans = , d;
    for (int i = ; i < m; ++i)
        ans += p[to][i];
    ans = all-ans;
    d = gcd(all, ans);
    if (ans == )
        printf("No solution\n");
    else
        printf("%d/%d\n", all/d, ans/d);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = ; times < T; ++times)
    {
        input();
        work();
    }
    return ;
}