[数位DP 高精度拓扑排序 bitset] BZOJ 2913 [Poi1997]XOR Gates

最新推荐文章于 2024-02-04 20:47:10 发布

原创最新推荐文章于 2024-02-04 20:47:10 发布 · 451 阅读

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本文介绍了一种结合拓扑排序与数位动态规划（数位DP）的方法来解决特定类型的问题。首先通过拓扑排序确定计算顺序，然后利用数位DP技巧计算可能的状态数目。文中提供了一个详细的代码实现案例。

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先拓扑再xor出输出的表达式

然后就是数位DP了

到底用不用高精度？

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<bitset>
using namespace std;

inline char nc(){
  static char buf[100000],*p1=buf,*p2=buf;
  if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
  return *p1++;
}

inline void read(int &x){
  char c=nc(),b=1;
  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

inline void read(int *x){
  char c=nc(); int len=0;
  for (;!(c>='0' && c<='1');c=nc());
  for (;c>='0' && c<='1';x[++len]=c-'0',c=nc());
}

const int con=100000000;
class Int{
public:long long a[10];
  void getdata(int x){memset(a,0,sizeof(a));while (x){a[++a[0]]=x%con;x=x/con;}}
  void pri(bool flag){
    if (a[0]==0||(a[0]==1&&a[1]==0)){printf("0");if (flag)printf("\n");return;}
    printf("%lld",a[a[0]]);
    for (int i=a[0]-1;i;i--)
      printf("%08lld",a[i]);
    if (flag)printf("\n");
  }
  bool operator <(const Int &X){
    if (a[0]<X.a[0])return true;if (a[0]>X.a[0])return false;
    for (int i=a[0];i;i--){if (a[i]<X.a[i])return true;if (a[i]>X.a[i])return false;}
    return false;
  }
  bool operator >(const Int &X){
    if (a[0]<X.a[0])return false;if (a[0]>X.a[0])return true;
    for (int i=a[0];i;i--){if (a[i]<X.a[i])return false;if (a[i]>X.a[i])return true;}
    return false;
  }
  bool operator <=(const Int &X){
    if (a[0]<X.a[0])return true;if (a[0]>X.a[0])return false;
    for (int i=a[0];i;i--){if (a[i]<X.a[i])return true;if (a[i]>X.a[i])return false;}
    return true;
  }
  bool operator >=(const Int &X){
    if (a[0]<X.a[0])return false;if (a[0]>X.a[0])return true;
    for (int i=a[0];i;i--){if (a[i]<X.a[i])return false;if (a[i]>X.a[i])return true;}
    return true;
  }
  bool operator ==(const Int &X){
    if (a[0]!=X.a[0])return false;for (int i=a[0];i;i--)if (a[i]!=X.a[i])return false;
    return true;
  }
  Int operator +(const Int &X){
    Int c;memset(c.a,0,sizeof(c.a));
    for (int i=1;i<=a[0]||i<=X.a[0];i++)
      {c.a[i]=c.a[i]+a[i]+X.a[i];c.a[i+1]+=c.a[i]/con;c.a[i]%=con;}
    c.a[0]=max(a[0],X.a[0]);if (c.a[c.a[0]+1])c.a[0]++;
    return c;
  }
  Int operator +(int num){
    Int c;memcpy(c.a,a,sizeof(c.a));c.a[1]+=num;
    for (int i=1;i<=c.a[0]&&c.a[i]>=con;i++)c.a[i]-=con,c.a[i+1]++;
    while (c.a[c.a[0]+1])c.a[0]++;
    return c;
  }
  Int operator -(const Int &X){
    Int c;memcpy(c.a,a,sizeof(c.a));
    for (int i=1;i<=a[0];i++){c.a[i]=c.a[i]-X.a[i];if (c.a[i]<0){c.a[i+1]--;c.a[i]+=con;}}
    while (c.a[0]&&!c.a[c.a[0]])c.a[0]--;
    return c;
  }
  Int operator -(int num){
    Int c;memcpy(c.a,a,sizeof(c.a));c.a[1]-=num;
    for (int i=1;i<=c.a[0]&&c.a[i]<0;i++)c.a[i]+=con,c.a[i+1]--;
    while (c.a[0]&&!c.a[c.a[0]])c.a[0]--;
    return c;
  }
  Int operator *(const Int &X){
    Int c;memset(c.a,0,sizeof(c.a));
    for (int i=1;i<=a[0];i++)for (int j=1;j<=X.a[0];j++)
                   {c.a[i+j-1]+=a[i]*X.a[j];c.a[i+j]+=c.a[i+j-1]/con;c.a[i+j-1]%=con;}
    c.a[0]=max(a[0]+X.a[0]-1,0ll);if (c.a[a[0]+X.a[0]]>0)c.a[0]++;
    return c;
  }
  Int operator *(int num){
    Int c;memset(c.a,0,sizeof(c.a));
    for (int i=1;i<=a[0];i++){c.a[i]+=a[i]*num;if (c.a[i]>=con){c.a[i+1]+=c.a[i]/con;c.a[i]%=con;}}
    c.a[0]=a[0];if (c.a[c.a[0]+1]>0)c.a[0]++;
    return c;
  }
  Int operator /(int num){
    Int c;memset(c.a,0,sizeof(c.a));
    long long x=0;for (int i=a[0];i;i--){x=x*con+a[i];c.a[i]=x/num;x=x%num;}
    c.a[0]=a[0];if (c.a[0]&&!c.a[c.a[0]])c.a[0]--;
    return c;
  }
};

const int N=4005;

struct edge{
  int u,v,next;
}G[N<<1];
int head[N],inum;

inline void add(int u,int v,int p){
  G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;
}

int lst[N],pnt;
int vst[N];
#define V G[p].v
inline void dfs(int u){
  vst[u]=1;
  for (int p=head[u];p;p=G[p].next)
    if (!vst[V])
      dfs(V);
  lst[++pnt]=u;
}

int n,m,T;
bitset<105> f[N];
int a[105];
Int F[105][2];
int s1[105],s2[105];

inline Int Solve(int *s){
  int cur=0;
  for (int i=1;i<=n;i++){
    F[i][0].getdata(0);
    F[i][1].getdata(0);
    if (s[i])
      F[i][cur]=F[i][cur]+1;
    for (int j=0;j<2;j++)
      for (int k=0;k<2;k++)
	F[i][j^(a[i]*k)]=F[i][j^(a[i]*k)]+F[i-1][j];
    cur^=a[i]*s[i];
  }
  return F[n][1];
}

int main(){
  int iu,iv;
  freopen("t.in","r",stdin);
  freopen("t.out","w",stdout);
  read(n); read(m); read(T);
  for (int i=1;i<=m;i++)
    read(iu),read(iv),iu=iu<0?-iu:iu+n,iv=iv<0?-iv:iv+n,add(iu,i+n,++inum),add(iv,i+n,++inum);
  for (int i=1;i<=n+m;i++) if (!vst[i]) dfs(i);
  reverse(lst+1,lst+pnt+1);
  for (int i=1;i<=n;i++) f[i][i]=1;
  for (int i=1;i<=n+m;i++)
    for (int p=head[lst[i]];p;p=G[p].next)
      f[V]^=f[lst[i]];
  for (int i=1;i<=n;i++) a[i]=f[n+T][i];
  read(s1); read(s2);
  Int Ans=(Solve(s2)-Solve(s1));
  int tem=0;
  for (int i=1;i<=n;i++) tem^=a[i]*s2[i];
  if (tem) Ans=Ans+1;
  Ans.pri(1);
  return 0;
    
}