HDU6321：Problem C. Dynamic Graph Matching（状压dp）

Problem Description

In the mathematical discipline of graph theory, a matching in a graph is a set of edges without common vertices.
You are given an undirected graph with n vertices, labeled by 1,2,...,n. Initially the graph has no edges.
There are 2 kinds of operations :
+ u v, add an edge (u,v) into the graph, multiple edges between same pair of vertices are allowed.
- u v, remove an edge (u,v), it is guaranteed that there are at least one such edge in the graph.
Your task is to compute the number of matchings with exactly k edges after each operation for k=1,2,3,...,n2. Note that multiple edges between same pair of vertices are considered different.

 

 

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there are 2 integers n,m(2≤n≤10,nmod2=0,1≤m≤30000), denoting the number of vertices and operations.
For the next m lines, each line describes an operation, and it is guaranteed that 1≤u

 

 

Output

For each operation, print a single line containing n2 integers, denoting the answer for k=1,2,3,...,n2. Since the answer may be very large, please print the answer modulo 109+7.

 

 

Sample Input

1 4 8 + 1 2 + 3 4 + 1 3 + 2 4 - 1 2 - 3 4 + 1 2 + 3 4

 

 

Sample Output

1 0 2 1 3 1 4 2 3 1 2 1 3 1 4 2

 

 

Source

2018 Multi-University Training Contest 3

题意：给一个图有N个点和若干个询问，每个询问增或者删一条边，每个询问输出选择1~N/2条不相交的边的方案数。

思路：因为点的个数很少，所以dp[i]表示i状态点集的选边方案数，因为X个点限定选X/2条边，所以很方便转移。

/*dp[i][j]表示j状态点集的方案数*/

# include 
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
int dp[2][1025], num[1025], sta[1025], ans[6];
void add(int &x, int y)
{
    x += y;
    if(x >= mod) x -= mod;
}
void sub(int &x, int y)
{
    x -= y;
    if(x < 0) x += mod;
}
int main()
{
    char c;
    int n, q, T, u, v, cnt=0;
    for(int i=0; i<1024; ++i)
    {
        num[i]= num[i>>1]+(i&1);//i的二进制有几个1
        if(~num[i]&1) sta[cnt++] = i;//sta保存偶数个1的二进制数
    }
    for(scanf("%d",&T);T;--T)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d%d",&n,&q);
        int now = 0;
        dp[0][0] = 1;
        while(q--)
        {
            getchar();
            c = getchar();
            scanf("%d%d",&u,&v);
            --u,--v;
            memset(ans, 0, sizeof(ans));
            int tmp = (1<