hdu 5305(爆搜+二染色) Friends

题意：

有n个人m对朋友,朋友有两种一种online的,一种face to face的。现在要让你对这些朋友性质进行安排,让每一个人的线上朋友和线下朋友人数是相同的,求满足的方案数。

思路

直接暴力染色。
然后减减枝。
一个就是对于每个人的度一定是偶数。
另外一个就是染色中,对于一个节点染色的不能超过一半才能继续dfs。

参考code:

/*
 #pragma warning (disable: 4786)
 #pragma comment (linker, "/STACK:0x800000")
 */
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <iterator>
#include <utility>
using namespace std;

template< class T > T _abs(T n)
{
    return (n < 0 ? -n : n);
}
template< class T > T _max(T a, T b)
{
    return (!(a < b) ? a : b);
}
template< class T > T _min(T a, T b)
{
    return (a < b ? a : b);
}
template< class T > T sq(T x)
{
    return x * x;
}
template< class T > T gcd(T a, T b)
{
    return (b != 0 ? gcd<T>(b, a%b) : a);
}
template< class T > T lcm(T a, T b)
{
    return (a / gcd<T>(a, b) * b);
}
template< class T > bool inside(T a, T b, T c)
{
    return a<=b && b<=c;
}


#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define F(i, n) for(int (i)=0;(i)<(n);++(i))
#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))
#define repok(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))
#define MEM0(addr) memset((addr), 0, sizeof((addr)))
#define MP(x, y) make_pair(x, y)
#define REV(s, e) reverse(s, e)
#define SET(p) memset(pair, -1, sizeof(p))
#define CLR(p) memset(p, 0, sizeof(p))
#define MEM(p, v) memset(p, v, sizeof(p))
#define CPY(d, s) memcpy(d, s, sizeof(s))
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
#define SZ(c) (int)c.size()
#define PB(x) push_back(x)
#define ff first
#define ss second
#define ll long long
#define ld long double
#define pii pair< int, int >
#define psi pair< string, int >
#define ls u << 1
#define rs u << 1 | 1
#define lson l, mid, u << 1
#define rson mid, r, u << 1 | 1

const int maxn = 30;
const int maxnum = 100010;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
using namespace std;
int m,n,ans;
int x[maxn],y[maxn],s[maxn],s1[maxn],s2[maxn],G[maxn][maxn];
void dfs(int dep){
    if(dep > m){
        rep(i,1,n){
            if(s1[i] != s2[i]) 
                return;
        }
        ans++; return;
    }
    int u = x[dep];
    int v = y[dep];
    if(s1[u] < s[u]/2 && s1[v] < s[v]/2){
        s1[u]++;s1[v]++;
        dfs(dep+1);
        s1[u]--;s1[v]--;
    }
    if(s2[u] < s[u]/2 && s2[v] < s[v]/2){
        s2[u]++;s2[v]++;
        dfs(dep+1);
        s2[u]--;s2[v]--;
    }
}
int main(){
    //READ("in.txt");
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        MEM0(G);
        rep(i,1,m){
            scanf("%d%d",&x[i],&y[i]);
            G[x[i]][y[i]] = G[y[i]][x[i]] = 1;
        }
        int flag = 1;
        rep(i,1,n){
            s[i] = s1[i] = s2[i] = 0;
            rep(j,1,n){
                if(i != j && G[i][j] == 1)
                    s[i]++;
            }
            if(s[i] & 1) flag = 0;
        }
        if(!flag){
            printf("0\n");
            continue;
        }
        ans = 0;
        dfs(1);
        printf("%d\n",ans);
    }
}