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

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].vinline 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;    }