这题主要是计算连通子图的个数(c)和不连通子图的个数(dc)还有连通度为1的子图的个数(c1)和连通度为2以上的子图的个数(c2)之间的转化关系
主要思路大概如下：
用状态压缩的方法算出状态为x的子图的不连通子图个数dc[x]，dc[x] = ∑ c[i]*(2^edge[x-i])，i为x的子集且i中有x的编号最小的元素，edge[x] 表示x集合内有几条边
连通子图个数c[x] = 2^edge[x] - dc[x]
想得到双连通子图的个数就要计算单连通子图的个数
单连通子图缩块后是一棵树，如果每次我们选择标号最小的点所在的块为根节点（块）
那么单连通子图可以看成是在这个双连通的根节点（块）的基础上连接一个连通分量，这样能枚举到所有的情况，也不会重复
mc[s][x] += mc[s][x - y] * c[y] * e[s][y]，其中mc[s][x-y]是指把x-y连接到s的方法数，e[s][y]是指s到y的边数
c1[s] += mc[x][s - x]，c1[s]是s中单连通子图的个数
而双连通子图个数 c2[s] = c[s] - c1[s]
最后转回去计算mc[s][0]，意思如果根节点s(块)不拓展连通分量的方法数，就相当于计算根节点（块）为双连通子图的方法数，等于c2[s]
再通过这些值计算mc[s+1][?]的值，不断的往上递推来完成全部的计算
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b) for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a) for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b) for( int i = (a) ; i >= (b) ; i --)
#define S64(a) scanf(in64,&a)
#define SS(a) scanf("%d",&a)
#define LL(a) ((a)<<1)
#define RR(a) (((a)<<1)+1)
#define pb push_back
#define pf push_front
#define X first
#define Y second
#define CL(Q) while(!Q.empty())Q.pop()
#define MM(name,what) memset(name,what,sizeof(name))
#define MC(a,b) memcpy(a,b,sizeof(b))
#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define read freopen("out.txt","r",stdin)
#define write freopen("out2.txt","w",stdout)
const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int mod = 1000000007;
const int maxn = 1 << 10;
int n, m;
int a[10][10];
i64 pow2[maxn];
i64 edge[maxn];
i64 ex[10][maxn];
i64 e[maxn][maxn];
i64 dc[maxn];
i64 c[maxn];
i64 c1[maxn];
i64 c2[maxn];
i64 mc[maxn][maxn];
vector<int>vx;
vector<int>v;
vector<int>v2;
void start(){
MM(edge, 0); MM(dc, 0); MM(c, 0); MM(c1, 0); MM(c2, 0); MM(e, 0); MM(ex, 0); MM(mc, 0);
for (int x = 0; x < n; x++){
for (int s = 0; s < (1 << n); s++){
for (int i = 0; i < n; i++)if (s&(1 << i)){
if (!a[x][i]){
ex[x][s] ++;
}
}
}
}
for (int s = 0; s < (1<<n); s++){
for (int x = 0; x < (1 << n); x++){
for (int i = 0; i < n; i++)if(s&(1<<i)){
e[s][x] += ex[i][x];
}
}
}
for (int s = 1;s < (1 << n); s++){
for (int i = 0; i < n; i++) if(s & (1<<i)){
for (int j = i + 1; j < n; j++) if(s& (1<<j)){
if (!a[i][j]){
edge[s]++;
}
}
}
}
int head;
for (int s = 1; s < (1 << n); s++){
v.clear();
for (int i = 0; i < n; i++){
if (s & (1 << i)){
v.push_back((1<<i));
}
}
head = v[0];
v.erase(v.begin());
int x;
for (int i = 0; i < (1 << ((int)v.size())); i++){
x = 0;
for (int pos = 0; pos < v.size(); pos++){
if (i &(1 << pos)) {
x += v[pos];
}
}
x += head;
if (x != s){
dc[s] += c[x] * pow2[edge[s - x]];
dc[s] %= mod;
}
}
c[s] = pow2[edge[s]] - dc[s] + mod;
c[s] %= mod;
}
for (int s = 1; s < (1 << n); s++){
vx.clear();
v.clear();
int x;
for (int i = 0; i < n; i++){
if (s & (1 << i)){
vx.push_back(1<<i);
}
}
int vxhead = vx[0];
vx.erase(vx.begin());
for (int i = 0; i < (1 << ((int)vx.size())); i++){
x = 0;
for (int pos = 0; pos < vx.size(); pos++){
if (i&(1 << pos)){
x += vx[pos];
}
}
x += vxhead;
if (x != s){
c1[s] += mc[x][s - x];
c1[s] %= mod;
}
}
c2[s] = (c[s] - c1[s]+mod)%mod;
mc[s][0] = c2[s];
for (int i = 0; i < n; i++){
if (s&(1 << i)){
head = i;
break;
}
}
for (int i = head + 1; i < n; i++){
if (!(s&(1 << i))){
v.push_back(1<<i);
}
}
for (int i = 1; i < (1 << ((int)v.size())); i++){
x = 0;
for (int pos = 0; pos < v.size(); pos++){
if (i&(1 << pos)){
x += v[pos];
}
}
v2.clear();
for (int u = 0; u < n; u++){
if (x&(1 << u)){
v2.push_back(1 << u);
}
}
int y;
for (int j = 1; j < (1 << ((int)v2.size())); j++) if(j&1){
y = 0;
for (int pos = 0; pos < v2.size(); pos++){
if (j & (1 << pos)){
y += v2[pos];
}
}
mc[s][x] +=(( mc[s][x - y] * c[y])%mod) * e[s][y];
mc[s][x] %= mod;
}
}
}
}
int main(){
pow2[0] = 1;
for (int i = 1; i < maxn; i++){
pow2[i] = pow2[i - 1] * 2;
pow2[i] %= mod;
}
int T;
cin >> T;
while (T--){
cin >> n >> m;
MM(a, 0);
int x, y;
for (int i = 0; i < n; i++){
a[i][i] = 1;
}
for (int i = 1; i <= m; i++){
cin >> x >> y;
x--; y--;
a[x][y] = a[y][x] = 1;
}
start();
cout << c2[(1 << n) - 1] << endl;
}
return 0;
}
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原文链接：hdu 4997 Biconnected，转载请注明来源！