soj4271 Love Me, Love My Permutation (DFS)

4271: Love Me, Love My Permutation

Description

Given a permutation of n: a[0], a[1] ... a[n-1], ( its elements range from 0 to n-1, For example: n=4, one of the permutation is 2310) we define the swap operation as: choose two number i, j ( 0 <= i < j <= n-1 ), take a[i] out and then insert a[i] to the back of a[j] of the initial permutation to get a new permutation :a[0], a[1], ..., a[i-1], a[i+1], a[i+2] ... a[j-1], a[j], a[i], a[j+1], ..., a[n-1]. For example: let n = 5 and the permutation be 03142, if we do the swap operation be choosing i = 1, j = 3, then we get a new permutation 01432, and if choosing i = 0, j = 4, we get 31420.

Given a confusion matrix of size n*n which only contains 0 and 1 (ie. each element of the matrix is either 0 or 1), the confusion value of a permutation a[0],a[1], ..., a[n-1] can be calculated by the following function:

int confusion()
{
	int result = 0;
	for(int i = 0;i < n-1; i++)
		for(int j = i+1; j < n; j++)
		{
			result = result + mat[a[i]][a[j]];
		}
	return result;
}

Besides, the confusion matrix satisfies mat[i][i] is 0 ( 0 <= i < n) and mat[i][j] + mat[j][i] = 1 ( 0 <= i < n, 0 <= j < n, i != j ) given the n, the confusion matrix mat, you task is to find out how many permutations of n satisfies: no matter how you do the swap function on the permutation(only do the swap function once), its confusion value does not increase.

Input

The first line is the number of cases T(1 <= T <= 5), then each case begins with a integer n (2 <= n <= 12), and the next n line forms the description of the confusion matrix (see the sample input).

Output

For each case , if there is no permutation which satisfies the condition, just print one line “-1”, or else, print two lines, the first line is a integer indicating the number of the permutations satisfy the condition, next line is the Lexicographic smallest permutation which satisfies the condition.

Sample Input

2
2
0 1
0 0
3
0 0 0
1 0 1
1 0 0

Sample Output

1
0 1
1
1 2 0

题解：DFS
   从后向前搜索，放进去的一位要和后面的每一段（n~n-1,n~n-2...n~0）保持1的个数大于0的个数，如果成立，则这一位可以放这个数，在向前继续搜索。
因为只能前面的数插入到后面，所以不需要考虑新加进去的数会不会对后面已经排好的短造成影响。也是这个原因要从后向前搜索。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#define M(a,b) memset(a,b,sizeof(a))
using namespace std;

int ans[25];
int num[25][25];
int out[25];
int n;
int ct;

void print()
{
    //cout<<'?'<<endl;
    int flag = 1;
    for(int i = n-1;i>=0;i--)
    {
        if(ans[i]<out[i]) {flag = 0; break;}
        if(ans[i]>out[i]) break;
    }
    if(!flag)
        for(int i = n-1;i>=0;i--)
           out[i] = ans[i];
    ct++;
}

void dfs(int pos)
{
    if(pos>=n)
    {
        print();
        return;
    }
    if(pos == 0)
    {
        for(int i = 0;i<n;i++)
        {
            ans[pos] = i;
            dfs(pos+1);
        }
    }
    else
    {
    for(int i = 0;i<n;i++)
    {
        int cnt = 0;
        int flag = 1;
        for(int j = pos-1;j>=0;j--)
        {

            if(num[i][ans[j]]==0) {flag = 0;break;}
            cnt += num[i][ans[j]];
            //cout<<i<<' '<<ans[j]<<' '<<num[i][ans[j]]<<' '<<cnt<<endl;
            if(cnt<0) {flag = 0;break;}
        }
        if(flag) {ans[pos] = i; dfs(pos+1);}
    }
    }
    return;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ct = 0;
        scanf("%d",&n);
        M(num,0);
        M(ans,0);
        M(out,0x3f);
        for(int i = 0;i<n;i++)
            for(int j = 0;j<n;j++)
        {
            scanf("%d",&num[i][j]);
            if(i == j) num[i][j] = 0;
            else if(num[i][j]==0)
                num[i][j] = -1;
        }
        dfs(0);
        if(ct==0) {puts("-1"); continue;}
        printf("%d\n",ct);
        for(int j = n-1;j>0;j--)
            printf("%d ",out[j]);
        printf("%d\n",out[0]);
    }
    return 0;
}