P2055 [ZJOI2009]假期的宿舍（二分图）

这题坑人的是题意，看了题解才搞明白

有一群学生和朋友，朋友会去看某些学生，互相认识的人之间可以互睡床位，问你能不能使得所有人都有床睡觉

人向床连边，如果是在校学生且不回家就向自己连一条边

然后统计非在校学生和在校学生且不回家的人数，最大匹配大于这个人数即OK

然后是多组记得清零

剩下跑一遍匈牙利即可
代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a) 
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=55;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
    char ch=getchar();int s=0,w=1;
    while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}
    while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
    return s*w;
}
inline void write(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+48);
}
int gcd(int a, int b){
    if(a==0) return b;
    if(b==0) return a;
    if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
    else if(!(b&1)) return gcd(a,b>>1);
    else if(!(a&1)) return gcd(a>>1,b);
    else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int _,n;
bool is[maxn],go[maxn],know[maxn][maxn];
int mp[maxn][maxn],lk[maxn],vis[maxn];

bool dfs(int x){
    re(y,1,n){
        if(mp[x][y]&&!vis[y]){
            vis[y]=1;
            if(!lk[y]||dfs(lk[y])){
                lk[y]=x;
                return 1;
            }
        }
    }
    return 0;
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>_;
    while(_--){
        cin>>n;
        re(i,1,n) cin>>is[i];
        re(i,1,n) cin>>go[i];
        re(i,1,n) re(j,1,n) cin>>know[i][j];
        mem0(mp),mem0(lk);
        int tot=0,ans=0;
        re(i,1,n)
            if(is[i]&&!go[i]) mp[i][i]=1;
        re(i,1,n){
            re(j,1,n){
                if(i==j) continue;
                if(is[j]&&know[i][j]) mp[i][j]=1;
            }
        }
        re(i,1,n) if(!is[i]||(is[i]&&!go[i])) ++tot;
        re(i,1,n){
            if(!is[i]||(is[i]&&!go[i])){
                mem0(vis);
                if(dfs(i)) ++ans;
            }
        }
        if(ans==tot) cout<<"^_^"<<endl;
        else cout<<"T_T"<<endl;
    }
    return 0;
}

标签：二分,typedef,return,int,re,maxn,P2055,ZJOI2009,define
来源： https://www.cnblogs.com/oneman233/p/11668303.html