acm_Sequence one

题目：

[align=left]Problem Description[/align]
Search is important in the acm algorithm. When you want to solve a problem by using the search method, try to cut is very important.<br>Now give you a number sequence, include n (<=1000) integers, each integer not bigger than 2^31,
you want to find the first P subsequences that is not decrease (if total subsequence W is smaller than P, than just give the first W subsequences). The order of subsequences is that: first order the length of the subsequence. Second order the sequence of each
integer’s position in the initial sequence. For example initial sequence 1 3 2 the total legal subsequences is 5. According to order is {1}; {3}; {2}; {1,3}; {1,2}. {1,3} is first than {1,2} because the sequence of each integer’s position in the initial sequence
are {1,2} and {1,3}. {1,2} is smaller than {1,3}. If you also can not understand , please see the sample carefully. <br>

[align=left]Input[/align]
The input contains multiple test cases.<br>Each test case include, first two integers n, P. (1<n<=1000, 1<p<=10000). <br>

[align=left]Output[/align]
For each test case output the sequences according to the problem description. And at the end of each case follow a empty line.

[align=left]Sample Input[/align]

3 5<br>1 3 2<br>3 6<br>1 3 2<br>4 100<br>1 2 3 2<br>

[align=left]Sample Output[/align]

1<br>3<br>2<br>1 3<br>1 2<br><br>1<br>3<br>2<br>1 3<br>1 2<br><br>1<br>2<br>3<br>1 2<br>1 3<br>2 3<br>2 2<br>1 2 3<br>1 2 2<br><br><br><div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div>Hint : You must make sure each subsequence in the subsequences is unique.</div>

题意：给你N个数，基本上有点求它的子集的感觉，因为他不能重复。。

想法：深搜加剪枝即可。。

代码：

#include<iostream>

#include<cstring>

using namespace std;

int d[1009];

int n,p;

int l,c;

struct name

{

int now,pos;

};

name w[10009];

bool flag;

bool check(int s,int e)

{

for(int i=s+1;i<e;i++)

{

if(d[i]==d[e])

return false;

}

return true;

}

void set(int length)

{

for(int i=0;i<length-1;i++)

cout<<w[i].now<<" ";

cout<<w[length-1].now<<endl;

}

void dfs(int dep,int pos)

{

if(c>=p)

return;

if(dep==l)

{

c++;

flag=true;

set(l);

return;

}

for(int i=pos;i<n;i++)

{

if((dep!=0&&w[dep-1].now<=d[i])||dep==0)

{

if(dep==0&&!check(-1,i))

continue;

if(dep!=0&&!check(w[dep-1].pos,i))

continue;

w[dep].now=d[i];

w[dep].pos=i;

dfs(dep+1,i+1);

}

}

return;

}

int main()

{

int i;

while(cin>>n>>p)

{

for(i=0;i<n;i++)

cin>>d[i];

c=0;

for(i=1;i<n;i++)

{

flag=false;

l=i;

dfs(0,0);

if(c>=p||!flag)

break;

}

cout<<endl;

}

return 0;

}

感想：好难。。