2016 Multi-University Training Contest 1 Necklace 环排+二分匹配

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5727

题意：由2*N颗宝石构成的环（阴阳宝石均为N颗且标号均从1~N） 之后给定M组 a,b;表示阳宝石a若和阴宝石b相邻会使得阳宝石变暗，问所构成的环中阳宝石变暗的最少数量？

其中（1<=N<=9, 1<= M <= N*N）

思路：确定一个阴宝石（我代码中让1不动），其余的环排。

将空隙变成一个点，先将所夹空隙的两点与阳宝石每点建图之后跑匈牙利即可，这跑出来的是最大匹配数（即最多不变暗的阳宝石数）；

稍加剪枝还是容易过的；8！ = 40320

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<bits/stdc++.h>
 using namespace std;
 #define rep0(i,l,r) for(int i = (l);i < (r);i++)
 #define rep1(i,l,r) for(int i = (l);i <= (r);i++)
 #define rep_0(i,r,l) for(int i = (r);i > (l);i--)
 #define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
 #define MS0(a) memset(a,0,sizeof(a))
 #define MS1(a) memset(a,-1,sizeof(a))
 #define MSi(a) memset(a,0x3f,sizeof(a))
 #define inf 0x3f3f3f3f
 #define A first
 #define B second
 #define MK make_pair
 #define esp 1e-8
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define bitnum(a) __builtin_popcount(a)
 #define clear0 (0xFFFFFFFE)
 #define mod 1000000007
 typedef pair<int,int> PII;
 typedef long long ll;
 typedef unsigned long long ull;
 template<typename T>
 void read1(T &m)
 {
     T x = ,f = ;char ch = getchar();
     while(ch <'' || ch >''){ if(ch == '-') f = -;ch=getchar(); }
     while(ch >= '' && ch <= ''){ x = x* + ch - '';ch = getchar(); }
     m = x*f;
 }
 template<typename T>
 void read2(T &a,T &b){read1(a);read1(b);}
 template<typename T>
 void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
 template<typename T>
 void out(T a)
 {
     if(a>) out(a/);
     putchar(a%+'');
 }
 inline ll gcd(ll a,ll b){ return b == ? a: gcd(b,a%b); }
 template<class T1, class T2> inline void gmin(T1& a,T2 b) { if(a > b) a = b; }
 template<class T1, class T2> inline void gmax(T1& a,T2 b) { if(a < b) a = b; }
 int S[][], f[], n;
 int vis[], girl[], line[][];
 int dfs(int p)
 {
     rep1(i,,n){
         if(line[p][i] && !vis[i]){
             vis[i] = ;
             if(!girl[i] || dfs(girl[i])){
                 girl[i] = p;
                 return ;
             }
         }
     }
     return ;
 }

 void init()
 {
     MS0(girl);MS0(line);
     rep1(i,,n) rep1(j,,n) if(!S[i][f[j]] && !S[i][f[j+]]) line[i][j] = ;
 }

 int solve()
 {
     init(); //建边
     int ans = ;
     rep1(i,,n){
         MS0(vis);
         if(dfs(i)) ans++;
     }
     return n-ans;
 }
 int main()
 {
     //freopen("data.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int m, a, b;
     while(scanf("%d%d",&n, &m) == ){
         MS0(S);
         rep1(i,,m){
             read2(a,b);
             S[a][b] = ;
         }
         rep1(i,,n) f[i] = i; f[n+] = ;
         int ans = inf;
         do{
             gmin(ans, solve());
         }while(next_permutation(f+,f+n+) && ans);
         printf("%d\n",ans);
     }
     return ;
 }