fzyzojP3618 -- [校内训练-互测20180412]士兵的游戏

二分图匈牙利也可以

判断必须点就看能不能通过偶数长度的增广路翻过去

代码：

（最后一个点4s多才行，，，卡不过算了）

开始边数写少了RE，应该是4*N*N

M-R随手开了一堆int？都要是long long

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define il inline
#define reg register int
#define numb (ch^'0')
using namespace std;
typedef long long ll;
il void rd(int &x){
    char ch;bool fl=false;
    while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
    for(x=numb;isdigit(ch=getchar());x=x*+numb);
    (fl==true)&&(x=-x);
}
namespace Miracle{
const int N=+;
ll a[N][N];
int co[N*N];
int in[N*N];
int n,m;
struct node{
    int nxt,to;
}e[*N*N];
int hd[N*N],cnt;
int num(int x,int y){
    return (x-)*m+y;
}
void add(int x,int y){
//    cout<<" ahahhaa "<<endl;
    e[++cnt].nxt=hd[x];
    e[cnt].to=y;
    hd[x]=cnt;
}
int to[N*N];
int vis[N*N];
bool dfs(int x,int id){
//    cout<<" x "<<x<<" id "<<id<<endl;
    for(reg i=hd[x];i;i=e[i].nxt){
        int y=e[i].to;
    //    cout<<" yy "<<y<<" vis "<<vis[y]<<endl;
        if(vis[y]==id) continue;
        vis[y]=id;
        if(!to[y]||dfs(to[y],id)){
            to[y]=x;
            in[x]=;
            to[x]=y;
            in[y]=;
            return true;
        }
    }
    return false;
}
bool fin(int x,int id,int pos){
    if(vis[x]==id) return false;
    vis[x]=id;
    if(pos==){
        if(!to[x]) return true;
        return fin(to[x],id,pos^);
    }
    for(reg i=hd[x];i;i=e[i].nxt){
        int y=e[i].to;
        if(y!=to[x]){
            if(fin(y,id,pos^)) return true;
        }
    }
    return false;
}
ll p[]={,,,,};
ll add(ll x,ll y,ll p){
    return x+y>=p?x+y-p:x+y;
}
ll qk(ll x,ll y,ll p){
    ll ret=;
    while(y){
        if(y&) ret=add(ret,x,p);
        x=add(x,x,p);
        y>>=;
    }
    return ret;
}
ll qm(ll x,ll y,ll p){
    ll ret=;
    while(y){
        if(y&) ret=qk(ret,x,p);
        x=qk(x,x,p);
        y>>=;
    }
    return ret;
}
bool che(ll x,ll d){
    int s=;
    ll lp=x-;
    while(!(lp&)){
        ++s;lp>>=;
    }
    ll now=qm(d,lp,x);
    if(now==||now==x-) return true;
    for(reg i=;i<s;++i){
        ll tmp=qk(now,now,x);
        if(tmp==&&(now!=&&now!=x-)) return false;
        now=tmp;
    }
    if(now!=) return false;
    return true;
}
bool M_R(ll x){
//    cout<<" xx "<<x<<endl;
    if(x==) return true;
    if(x==) return true;
    if(x==46856248255981ll) return false;
    if(x==||x==||x==||x==||x==||x==||x==||x==||x==||x==||x==) return true;
    for(reg i=;i<;++i){
        if(x%p[i]==) return false;
        if(!che(x,p[i])) return false;
    }
//    cout<<" ok "<<endl;
    return true;
}
int mem[N*N],nb;
int main(){
//    cout<<" M_R "<<M_R(1557403521852231)<<endl;
    rd(n);rd(m);
//    cout<<" 2333 "<<endl;
    for(reg i=;i<=n;++i){
        for(reg j=;j<=m;++j){
            scanf("%lld",&a[i][j]);
            co[num(i,j)]=(i+j)%;
        }
    }

//    return 0;
    for(reg i=;i<=n;++i){
        for(reg j=;j<=m;++j){
            if(a[i][j]==-) continue;
            if(j!=m){
                if(a[i][j+]!=-){
                    if(M_R(a[i][j]+a[i][j+])==){
                        add(num(i,j),num(i,j+));
                        add(num(i,j+),num(i,j));
                    }
                }
            }
            if(i!=n){
                if(a[i+][j]!=-){
                    if(M_R(a[i][j]+a[i+][j])==){
                        add(num(i,j),num(i+,j));
                        add(num(i+,j),num(i,j));
                //        cout<<num(i,j)<<" "<<num(i+1,j)<<endl;
                    }
                }
            }
        }
    }
//    cout<<" after build "<<endl;
    int tot=;
    int le=,ri=;
    for(reg i=;i<=n;++i){
        for(reg j=;j<=m;++j){
            if(co[num(i,j)]==&&a[i][j]!=-){
                ++le;
                tot+=dfs(num(i,j),num(i,j));
            }else if(a[i][j]!=-) ++ri;
        }
    }
//    cout<<" le ri "<<le<<" "<<ri<<" tot "<<tot<<endl;
    if(le==ri&&tot==le){
        puts("");return ;
    }
    memset(vis,,sizeof vis);
    for(reg i=;i<=n;++i){
        for(reg j=;j<=m;++j){
            if(a[i][j]!=-){
                if(!in[num(i,j)]||fin(to[num(i,j)],num(i,j),)){
                    mem[++nb]=num(i,j);
                }
            }
        }
    }
    sort(mem+,mem+nb+);
    printf("%d\n",nb);
    for(reg i=;i<=nb;++i){
        printf("%d %d\n",(mem[i]-)/m+,mem[i]-((mem[i]-)/m)*m);
    }
    return ;
}

}
signed main(){
//    freopen("data3618.in","r",stdin);
//    freopen("data3618.out","w",stdout);
    Miracle::main();
    return ;
}

/*
   Author: *Miracle*
   Date: 2019/2/2 20:45:23
*/

这个题的二分图匹配思想还是很巧妙

从最大匹配来考虑，便于决策