hdu 6321 hdu多校第三场c题（dp）
Problem C. Dynamic Graph Matching
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 180 Accepted Submission(s): 66
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
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思路： dp[ now ][ s ] 表示当前操作后， s 这个点集中 已经匹配的方案数。
那么对如果一个状态 s 这个点集 中没有 u 和 v 这两个点，并且现在要加入u和v的连边，那么我就可以通过s 这个状态转移到aims这个状态，aims 这个状态也就是从 s 和 u，v 点集的并。
同理如果我现在要删除u v这个点集 ，那么对于aims （包括 u v 的点集） 我们可以找到一个状态 i i 中没有u v两点，我们删除从i转移到aims 的方案数。
代码：
#include
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const int N =2005;
ll dp[2][N];
ll yi[40];
ll ans[15];
int cnt[N];
void init_yi()
{
yi[0]=1;
for(int i=1;i<=33;i++) yi[i]=yi[i-1]*2;
}
void init_cnt()
{
for(int i=0;i