Knapsack Cryptosystem（状压dp）

Title Description

Amy asks Mr. B problem E. Please help Mr. B to solve the following problem.

There are n people, who don't know each other at the beginning.

There are m turns. In each turn, 2 of them will make friends with each other.

The friend relation is mutual and transitive.

If A is a friend of B, then B is also a friend of A.

For example, if A is a friend of B, B is a friend of C, then A and C are friends.

At the beginning and after each turn, please calculate the number of ways to select four people from, such that any two of these four are not friends.

Enter a description:

The first line contains two integers, n and m (n <= 100000, m <= 200000), which are the number of people, and the number of turns.

In the following m lines, the i-th line contains two integers x and y ( 1 <= x <= n, 1 <= y <= n, x ≠ y), which means the x-th person and the y-th person make friends in the i-th turn.

The x-th person and y-th person might make friends in several turns.

Output Description:

Output m+1 lines, each line contains an integer, which is the number of quadruples.

Output at the beginning and after each turn, so there are m+1 lines.

Example 1

Entry

copy

Export

copy

Example 2

Entry

copy

100000 0

Export

copy

4166416671249975000

Explanation

Do not use int. 

Problem Solving Report: This is starting to look a title in a backpack, but then, backpack too much capacity, and there is no way to store an array, then you can consider like pressure, because it is less than n is equal to 36, 
is obviously no way to direct the compression-shaped, so it can be segmented to form a binary voltage, such complexity can be reduced jumping type, need to open a map stored in the latter half of the solution process to to query the map, and then 
each went to judge, to note is the need to set up two storage arrays, only one word can not get the desired result. 


ac Code:

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<map>
 7 using namespace std;
 8 typedef long long ll;
 9 
10 const int maxn = 1e6+1000;
11 
12 ll num1[maxn];
13 ll num2[maxn];
14 ll sum[maxn];
15 ll sum2[maxn];
16 int n;
17 ll s;
18 map<ll ,int> mp;
19 int main()
20 {
21     cin>>n>>s;
22     for(int i=1;i<=n;i++)
23         cin>>num1[i];
24     int mid=n/2;
25     for(int i=1;i<=mid;i++)
26         num2[i-1]=num1[i];
27     for(int i=0;i<(1<<mid);i++)
28     {
29         for(int j=0;j<mid;j++)
30         {
31             if(i&(1<<j))
32                 sum[i]+=num2[j];    
33         }
34         mp[sum[i]]=i;
35     }    
36     int mid2=n-mid;
37     for(int i=mid+1;i<=n;i++)
38         num2[i-mid-1]=num1[i];
39     for(int i=0;i<(1<<mid2);i++)
40     {
41         for(int j=0;j<mid2;j++)
42         {
43             if(i&(1<<j))
44                 sum2[i]+=num2[j];
45         }
46         if(mp.count(s-sum2[i]))
47         {
48             int tmp=mp[s-sum2[i]];
49             for(int j=0;j<mid;j++)
50             {
51                 if(tmp&(1<<j))
52                     cout<<1;
53                 else
54                     cout<<0; 
55             }
56             for(int j=0;j<mid2;j++)
57             {
58                 if(i&(1<<j))
59                     cout<<1;
60                 else
61                     cout<<0;
62             }
63             cout<<endl;
64         }
65     }
66 }