这状压dp还是比较好的。。思维上有点难度。。。

目标是使该点连的黑/白边数为度数的一半。。考虑到度数不超过8，那么边数自然也不超过4，所以只要用2位二进制来表示连的黑边数。。

然后转移就可以随便转移。。主要是在二进制部分位的提取和替换。。

/** *　　　　　　 　　┏┓　　 　┏┓ * 　　　　　 　　┏┛┗━━━━━━━┛┗━━━┓ * 　　　　　 　　┃　　　　　　　┃ 　 * 　　　　　 　　┃　　　━　　 　┃ * 　　　　　 　　┃　＞　　　＜　┃ * 　　　　　 　　┃　　　　　　　┃ * 　　　　　 　　┃...　⌒　... 　┃ * 　　　　　　 　┃ ┃ * 　　　　　　 　┗━┓ ┏━┛ * 　　　　　　　　　┃ ┃　Code is far away from bug with the animal protecting　　　　　　　　　　 * 　　　　　　　　　┃ ┃ 神兽保佑,代码无bug * 　　　　　　　　　┃ ┃　　　　　　　　　　　 * 　　　　　　　　　┃ ┃ 　　　　　　 * 　　　　　　　　　┃ ┃ * 　　　　　　　　　┃ ┃　　　　　　　　　　　 * 　　　　　　　　　┃ ┗━━━┓ * 　　　　　　　　　┃ ┣┓ * 　　　　　　　　　┃ ┏┛ * 　　　　　　　　　┗┓┓┏━━━━━━━━┳┓┏┛ * 　　　　　　　　　　┃┫┫ ┃┫┫ * 　　　　　　　　　　┗┻┛ ┗┻┛ */ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue> #include<cmath> #include<map> #include<stack> #include<set> #define inc(i,l,r) for(int i=l;i<=r;i++) #define dec(i,l,r) for(int i=l;i>=r;i--) #define link(x) for(edge *j=h[x];j;j=j->next) #define mem(a) memset(a,0,sizeof(a)) #define ll long long #define eps 1e-12 #define succ(x) (1<<x) #define lowbit(x) (x&(-x)) #define sqr(x) ((x)*(x)) #define mid (x+y>>1) #define NM 205 #define nm 100005 #define pi 3.1415926535897931 const ll inf=1000000007; using namespace std; ll read(){ ll x=0,f=1;char ch=getchar(); while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); return f*x; } int n,m,c[nm],d[nm],_x,_y,s,tot,b[NM]; int main(){ int _=read();while(_--){ n=read();m=read();mem(c);mem(b);mem(d); tot=succ(n*2)-1;s=0; c[0]=1; inc(i,1,m){ _x=read();_y=read();b[_x]++;b[_y]++; inc(j,0,tot)if(c[j]){ int x=2*!!(j&succ(2*_x-1))+!!(j&succ(2*_x-2)); int y=2*!!(j&succ(2*_y-1))+!!(j&succ(2*_y-2)); int dx=(j&succ(2*_x-1))+(j&succ(2*_x-2)); int dy=(j&succ(2*_y-1))+(j&succ(2*_y-2)); //printf("%d %d %d %d %d\n",j,dx,dy,x,y); x++;y++; if(x<5&&y<5)d[j^dx^dy^(x*succ(2*_x-2))^(y*succ(2*_y-2))]+=c[j]; } inc(j,0,tot)c[j]=d[j]+c[j],d[j]=0; } bool f=false; inc(i,1,n)if(b[i]%2){f=true;break;}else s+=(b[i]/2)*succ(i*2-2); if(!f)printf("%d\n",c[s]);else printf("0\n"); } return 0; }

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2705    Accepted Submission(s): 1299 Problem Description There are n people and m pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people wants to have the same number of online and offline friends (i.e. If one person has x onine friends, he or she must have x offline friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements.   Input The first line of the input is a single integer T (T=100), indicating the number of testcases. For each testcase, the first line contains two integers n (1≤n≤8) and m (0≤m≤n(n−1)2), indicating the number of people and the number of pairs of friends, respectively. Each of the next m lines contains two numbers x and y, which mean x and y are friends. It is guaranteed that x≠y and every friend relationship will appear at most once.   Output For each testcase, print one number indicating the answer.   Sample Input 2 3 3 1 2 2 3 3 1 4 4 1 2 2 3 3 4 4 1   Sample Output 0 2   Author XJZX   Source 2015 Multi-University Training Contest 2   Recommend wange2014   |   We have carefully selected several similar problems for you:   6297  6296  6295  6294  6293    Statistic |  Submit |  Discuss | Note