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;
}

  

 

soj4271 Love Me, Love My Permutation (DFS)的更多相关文章

  1. Tree and Permutation dfs hdu 6446

    Problem Description There are N vertices connected by N−1 edges, each edge has its own length.The se ...

  2. codeforces 691D D. Swaps in Permutation(dfs)

    题目链接: D. Swaps in Permutation time limit per test 5 seconds memory limit per test 256 megabytes inpu ...

  3. codeforces 691D Swaps in Permutation DFS

    这个题刚开始我以为是每个交换只能用一次,然后一共m次操作 结果这个题的意思是操作数目不限,每个交换也可以无限次 所以可以交换的两个位置连边,只要两个位置连通,就可以呼唤 然后连通块内排序就好了 #in ...

  4. HNU Joke with permutation (深搜dfs)

    题目链接:http://acm.hnu.cn/online/?action=problem&type=show&id=13341&courseid=0 Joke with pe ...

  5. 2018中国大学生程序设计竞赛 - 网络选拔赛 1009 - Tree and Permutation 【dfs+树上两点距离和】

    Tree and Permutation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Oth ...

  6. hdu6446 Tree and Permutation 2018ccpc网络赛 思维+dfs

    题目传送门 题目描述:给出一颗树,每条边都有权值,然后列出一个n的全排列,对于所有的全排列,比如1 2 3 4这样一个排列,要算出1到2的树上距离加2到3的树上距离加3到4的树上距离,这个和就是一个排 ...

  7. codeforces 285 D. Permutation Sum 状压 dfs打表

    题意: 如果有2个排列a,b,定义序列c为: c[i] = (a[i] + b[i] - 2) % n + 1 但是,明显c不一定是一个排列 现在,给出排列的长度n (1 <= n <= ...

  8. leetcode 784. Letter Case Permutation——所有BFS和DFS的题目本质上都可以抽象为tree,这样方便你写代码

    Given a string S, we can transform every letter individually to be lowercase or uppercase to create ...

  9. 【HDOJ6628】permutation 1(dfs)

    题意:求1到n的排列中使得其差分序列的字典序为第k大的原排列 n<=20,k<=1e4 思路:爆搜差分序列,dfs时候用上界和下界剪枝 #include<bits/stdc++.h& ...

随机推荐

  1. Tomcat热部署方法(3种)【转】

    热部署是指在你修改项目BUG的时候对JSP或JAVA类进行了修改在不重启WEB服务器前提下能让修改生效.但是对配置文件的修改除外! 1.直接把项目web文件夹放在webapps里. 2.在tomcat ...

  2. test md

    [TOC] Glossary SUT SYSTEM UNDER TEST CUT CLASS UNDER TEST MUT METHOD UNDER TEST Tests without Use of ...

  3. day1(变量、常量、注释、用户输入、数据类型)

    一.变量 name = "SmallNine" 等号前面是变量名(标识符),等号后面是变量值 变量的主要作用:就是把程序运算的中间结果临时存到内存里,已备后面的代码继续调用. 变量 ...

  4. 洛谷P1593 因子和

    题目描述 输入两个正整数a和b,求a^b的因子和.结果太大,只要输出它对9901的余数. 输入输出格式 输入格式: 仅一行,为两个正整数a和b(0≤a,b≤50000000). 输出格式: a^b的因 ...

  5. 队列Queue和栈

    1.队列Queue是常用的数据结构,可以将队列看成特殊的线性表,队列限制了对线性表的访问方式,只能从线性表的一段添加(offer)元素, 从另一段取出(poll)元素,队列遵循先进先出的原则. 2.J ...

  6. leetcode — interleaving-string

    /** * Source : https://oj.leetcode.com/problems/interleaving-string/ * * * Given s1, s2, s3, find wh ...

  7. JavaScript中常见的10个BUG及其修复方法

    如今网站几乎100%使用JavaScript.JavaScript看上去是一门十分简单的语言,然而事实并不如此.它有很多容易被弄错的细节,一不注意就导致BUG. 1. 错误的对this进行引用 在闭包 ...

  8. php分享二十七:批量插入mysql

    一:思考 1:如果插入的某个字段大于数据库定义的长度了,数据库会怎么处理? 1>如果数据库引擎是myisam,则数据库会截断后插入,不报错 2>如果数据库引擎是innodb,则数据库会报 ...

  9. Thinkphp动态切换主题

    'DEFAULT_THEME' => '2014', 'TMPL_DETECT_THEME' => true, // 自动侦测模板主题 'THEME_LIST' => '2012,2 ...

  10. Java门派的风险

    Java门派的风险 正在看周思博(www.joelonsoftware.com)的新文章.这次是疯狂攻击Java.主要论点是:Java不够难,作为工业语言不错,但作为学校的教学语言,就忒差了.学校应该 ...