HDU 6321 Problem C. Dynamic Graph Matching （2018多校3） 状压DP

Problem C. Dynamic Graph Matching

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1341    Accepted Submission(s): 561

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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题意：在一个无向图中给出 n 个点， n <= 10 且 n 为偶数，有m个操作，有两种操作，一种是在两个点之间增加一条边，一种是从图中去除两个点之间的一条边，可以有重边

每次操作都要求 n / 2 个答案  记为询问 X  (1...n/2) ， X = 2时表示：从图中取两条边，两条边中没有边共点的方案数，以此类推

思路：因为 n 只有十，考虑状态压缩， 1 << 10 = 1024 ，即最多需要 1024个十进制的数可以表示出所有点集的情况

 

例如样例中有 4 个点

插入 + 1 2 边的时候用二进制  _ _ 1 1 表示，即 3，而四个点的图中，包含了这两个点的点集且只包含偶数个点的情况只有两种

1. _ _ 1 1  = 3       2. 1 1 1 1  = 15  （本题点集含义：点数 = 2 （例如 点 1 和 点 2） 的点集 dp[3] 表示，有这两个在内，且符合题目要求的没有两边共点的方案数）

注意预处理的时候不可最后再取模，会爆int，所以最好一边算一边取，且用 long long 保证不丢失

代码：

#include
using namespace std;
#define ll long long
#define mem(a,x) memset(a,x,sizeof(a))

const int mod = 1e9 + 7;
const int maxn = 1 << 10 | 1;
int dp[2][maxn];
int bits[maxn],eve[maxn],ans[10];
int cnt;

void init(){
    int len = 1 << 10;
    cnt = 0;
    for(int i = 0;i < len;i++){
        bits[i] = bits[i >> 1] + (i & 1);    //  bits[i] 表示 i 的二进制有多少个1 
        if(bits[i] % 2 == 0){    
            eve[cnt++] = i;                // eve[i] 记录的是 1 的个数为偶数的二进制数
        }
    }
}

int main(){
    int T;
    int n,m,u,v;
    char opt[5];
    init();
    scanf("%d",&T);
    while(T--){
        mem(dp,0);
        dp[0][0] = 1;
        int flag = 0;             // 用 flag ^ 1 做滚动数组 
        scanf("%d %d",&n,&m);
        int len = 1 << n;
        while(m--){
            mem(ans,0);
            scanf("%s %d %d",opt,&u,&v);
            u--,v--;
            int state = (1 << u) | (1 << v);  // state 表示当前插入的边用二进制表示的状态 
            if(opt[0] == '+'){
                for(int i = 0;i < cnt;i++){
                    if(eve[i] >= len){
                        break;
                    }
                    int sta = eve[i];   //  
                    dp[flag ^ 1][sta] = dp[flag][sta];
                    if((sta & state) == state){
                        dp[flag ^ 1][sta] = (dp[flag ^ 1][sta] + dp[flag][sta ^ state]) % mod;
                    }
                    ans[bits[sta] / 2] = (ans[bits[sta] / 2] + dp[flag ^ 1][sta]) % mod;
                }
            }else{
                for(int i = 0;i < cnt;i++){
                    if(eve[i] >= len){
                        break;
                    }
                    int sta = eve[i];
                    dp[flag ^ 1][sta] = dp[flag][sta];
                    if((sta & state) == state){
                        dp[flag ^ 1][sta] -= dp[flag][sta ^ state];
                        if(dp[flag ^ 1][sta] < 0){
                            dp[flag ^ 1][sta] += mod;
                        }
                    }
                    ans[bits[sta] / 2] = (ans[bits[sta] / 2] + dp[flag ^ 1][sta]) % mod;
                }
            }
            flag ^= 1;
            for(int i = 1;i <= n / 2;i++){
                if(i != 1)
                    printf(" ");
                printf("%d",ans[i]);
            }
            puts("");
        }
    }
    return 0;
}