CodeChef KnightMov

码死了...考试的时候基本上是写一会儿思考一会儿人生....考完了调了调...最后400行+....不应该这么长的....以后重写一下再补题解.....

也许这就是蒟蒻吧.jpg

安利cstdio博客翻译的题解.

#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
using namespace std;
const int mod=1000000007;
int X,Y,K;
int ax,ay,bx,by;
int xx[20],yy[20];
set<pair<int,int> > dict;
double direc(int a,int b){
  if(a!=0)return double(b)/a;
  else return 1e8;
}
int check1(int a,int b){
  if(X==0&&Y==0)return -1;
  int step1,step2;
  if(a==0){
    if(X==0){
      double tmp=Y/(double)(b);
      if(tmp>0&&(fabs(tmp-floor(tmp))<1e-8)){
	step1=tmp;
	int tmpx=0,tmpy=0;
	for(int i=1;i<=step1;++i){
	  tmpx+=a;tmpy+=b;
	  if(dict.find(pair<int,int>(tmpx,tmpy))!=dict.end())return 0;
	}
	return -1;
	//step1=floor(tmp);
      }else return 0;
    }else return 0;
  }else{
    step1=X/(double)(a);
    if(step1<=0||(fabs(step1-floor(step1))>1e-8))return 0;
  }
  if(b==0){
    if(Y==0){
      double tmp=X/(double)(a);
      if(tmp>0&&(fabs(tmp-floor(tmp))<1e-8)){
	step1=tmp;
	int tmpx=0,tmpy=0;
	for(int i=1;i<=step1;++i){
	  tmpx+=a;tmpy+=b;
	  if(dict.find(pair<int,int>(tmpx,tmpy))!=dict.end())return 0;
	}
	return -1;
      }else return 0;
    }else return 0;
  }else{
    step2=Y/(double)(b);
    if(step2<=0||(fabs(step2-floor(step2))>1e-8))return 0;
  }
  if(step1==step2){
    //blabla
    int tmpx=0,tmpy=0;
    for(int i=1;i<=step1;++i){
      tmpx+=a;tmpy+=b;
      if(dict.find(pair<int,int>(tmpx,tmpy))!=dict.end())return 0;
    }
    return -1;
  }
  else{
    return 0;
  }
}
int qpow(int a,int x){
  int ans=1;
  for(;x;x>>=1,a=a*1ll*a%mod){
    if(x&1)ans=ans*1ll*a%mod;
  }
  return ans;
}
int check4(int a,int b){
  if(X==0&&Y==0)return -1;
  int step1,step2;
  if(a==0){
    if(X==0){
      double tmp=Y/(double)(b);
      if(tmp>0&&(fabs(tmp-floor(tmp))<1e-8)){
	step1=tmp;
	int tmpx=0,tmpy=0;
	for(int i=1;i<=step1;++i){
	  tmpx+=a;tmpy+=b;
	  if(dict.find(pair<int,int>(tmpx,tmpy))!=dict.end())return 0;
	}
	return 1;
	//step1=floor(tmp);
      }else return 0;
    }else return 0;
  }else{
    step1=X/(double)(a);
    if(step1<=0||(fabs(step1-floor(step1))>1e-8))return 0;
  }
  if(b==0){
    if(Y==0){
      double tmp=X/(double)(a);
      if(tmp>0&&(fabs(tmp-floor(tmp))<1e-8)){
	step1=tmp;
	int tmpx=0,tmpy=0;
	for(int i=1;i<=step1;++i){
	  tmpx+=a;tmpy+=b;
	  if(dict.find(pair<int,int>(tmpx,tmpy))!=dict.end())return 0;
	}
	return 1;
      }else return 0;
    }else return 0;
  }else{
    step2=Y/(double)(b);
    if(step2<=0||(fabs(step2-floor(step2))>1e-8))return 0;
  }
  if(step1==step2){
    //blabla
    int tmpx=0,tmpy=0;
    for(int i=1;i<=step1;++i){
      tmpx+=a;tmpy+=b;
      if(dict.find(pair<int,int>(tmpx,tmpy))!=dict.end())return 0;
    }
    return 1;
  }
  else{
    return 0;
  }
}
int gcd(int a,int b){
  return b==0?a:gcd(b,a%b);
}
int F[600],FF[600][600];
int seq[100],cnt=0;
int BASE=250000;
int q[500005];
bool dfn[500005];
bool block[500005];
bool legal(int x){
  return x>=0&&x<500005;
}
int sum1[500005],sum2[500005];
int cir[500005],T;
bool dfn2[500005];
int fff[500005];
int check3(int a,int b,int to){
  b=-b;
  memset(cir,0,sizeof(cir));
  memset(dfn,0,sizeof(dfn));
  memset(block,0,sizeof(block));
  memset(sum1,0,sizeof(sum1));memset(sum2,0,sizeof(sum2));
  for(int i=1;i<=cnt;++i)block[seq[i]+BASE]=true;
  for(int i=1;i<500005;++i){
    if(i>=a){
      sum1[i]=sum1[i-a]+(block[i]);
    }
    else{
      sum1[i]=(block[i]);
    }
    if(i>=b){
      sum2[i]=sum2[i-b]+(block[i]);
    }else{
      sum2[i]=(block[i]);
    }
  }
  ++T;
  for(int i=a*b+a+b;i<500005;++i){
    if(sum1[i]-sum1[i-a*b-a]==0&&sum2[i]-sum2[i-a*b-b]==0)cir[i]=true;
  }
  int head=0,tail=0;dfn[BASE]=true;
  q[tail++]=BASE;
  while(head!=tail){
    int x=q[head++];
    if(legal(x+a)&&!dfn[x+a]&&!block[x+a]){
      dfn[x+a]=true;q[tail++]=x+a;
    }
    if(legal(x-b)&&!dfn[x-b]&&!block[x-b]){
      dfn[x-b]=true;q[tail++]=x-b;
    }
  }
  if(!dfn[BASE+to])return 0;
  memset(dfn2,0,sizeof(dfn2));
  head=tail=0;dfn2[BASE+to]=true;
  q[tail++]=BASE+to;
  while(head!=tail){
    int x=q[head++];
    if(legal(x-a)&&!dfn2[x-a]&&!block[x-a]){
      dfn2[x-a]=true;q[tail++]=x-a;
    }
    if(legal(x+b)&&!dfn2[x+b]&&!block[x+b]){
      dfn2[x+b]=true;q[tail++]=x+b;
    }
  }
  for(int i=0;i<500005;++i){
    if(cir[i]&&dfn[i]&&dfn2[i]){
      return -1;
    }
  }memset(dfn,0,sizeof(dfn));
  memset(fff,0,sizeof(fff));fff[BASE]=1;
  head=0,tail=0;dfn[BASE]=true;
  q[tail++]=BASE;
  while(head!=tail){
    int x=q[head++];
    if(legal(x+a)&&!dfn[x+a]&&!block[x+a]){
      fff[x+a]=(fff[x+a]+fff[x])%mod;
      dfn[x+a]=true;q[tail++]=x+a;
    }
    if(legal(x-b)&&!dfn[x-b]&&!block[x-b]){
      fff[x-b]=(fff[x-b]+fff[x])%mod;
      dfn[x-b]=true;q[tail++]=x-b;
    }
  }
  return fff[BASE+to];
}
int check2(){//方向相同,可以同向,也可以反向
  if(ax==0){
    if(X!=0)return 0;
    if((ay>0)==(by>0)){
      //同向
      int gy=gcd(ay,by);
      memset(F,0,sizeof(F));
      if(Y%gy!=0||Y/gy<0)return 0;
      int A=ay/gy,B=by/gy;
      F[0]=1;
      for(int i=0;i<=Y/gy;++i){
	if(dict.find(pair<int,int>(0,gy*i))!=dict.end())F[i]=0;
	F[i+A]=(F[i+A]+F[i])%mod;F[i+B]=(F[i+B]+F[i])%mod;
      }
      return F[Y/gy];
    }else{
      //反向
      int gy=gcd(fabs(ay),fabs(by));
      if(Y%gy!=0)return 0;
      cnt=0;
      for(int i=1;i<=K;++i){
	if(xx[i]==0&&yy[i]%gy==0){
	  seq[++cnt]=yy[i]/gy;
	}
      }
      int A=ay/gy,B=by/gy;
      if(A>0)return check3(A,B,Y/gy);
      else return check3(B,A,Y/gy);
    }
  }else if(ay==0){
    if(Y!=0)return 0;
    if((ax>0)==(bx>0)){
      //同向
      int gx=gcd(ax,bx);
      memset(F,0,sizeof(F));
      if(X%gx!=0||X/gx<0)return 0;
      int A=ax/gx,B=bx/gx;
      F[0]=1;
      for(int i=0;i<=X/gx;++i){
	if(dict.find(pair<int,int>(gx*i,0))!=dict.end()){
	  F[i]=0;
	}
	F[i+A]=(F[i+A]+F[i])%mod;F[i+B]=(F[i+B]+F[i])%mod;
      }
      return F[X/gx];
    }else{
      //反向
      int gx=gcd(fabs(ax),fabs(bx));
      if(X%gx!=0)return 0;
      cnt=0;
      for(int i=1;i<=K;++i){
	if(yy[i]==0&&xx[i]%gx==0){
	  seq[++cnt]=xx[i]/gx;
	}
      }
      int A=ax/gx,B=bx/gx;
      if(A>0)return check3(A,B,X/gx);
      else return check3(B,A,X/gx);
    }
  }else{
    if((ax>0)==(bx>0)){
      //同向
      int gx=gcd(ax,bx),gy=gcd(ay,by);
      if(X%gx!=0||X/gx<0)return 0;
      if(Y%gy!=0||Y/gy<0)return 0;
      if(X/gx!=Y/gy)return 0;
      int A=ax/gx,B=bx/gx;
      memset(F,0,sizeof(F));
      F[0]=1;
      F[0]=1;
      for(int i=0;i<=X/gx;++i){
	if(dict.find(pair<int,int>(gx*i,gy*i))!=dict.end()){
	  F[i]=0;
	}
	F[i+A]=(F[i+A]+F[i])%mod;F[i+B]=(F[i+B]+F[i])%mod;
      }
      return F[X/gx];
    }else{
      //反向
      int gx=gcd(fabs(ax),fabs(bx)),gy=gcd(fabs(ay),fabs(by));
      if(X%gx!=0)return 0;
      if(Y%gy!=0)return 0;
      if(X/gx!=Y/gy)return 0;
      cnt=0;
      int A=ax/gx,B=bx/gx;
      for(int i=1;i<=K;++i){
	if(yy[i]%gy==0&&xx[i]%gx==0&&yy[i]/gy==xx[i]/gx){
	  seq[++cnt]=xx[i]/gx;
	}
      }
      if(A>0)return check3(A,B,X/gx);
      else return check3(B,A,X/gx);
    }
  }
  //同向:一定无环,递推一波即可

  //反向:可能有环...
  
}
int fac[500005],inv[500005];
void init(){
  fac[0]=1;
  for(int i=1;i<500005;++i)fac[i]=fac[i-1]*1ll*i%mod;
  inv[0]=1;
  inv[1]=1;
  for(int i=2;i<500005;++i)inv[i]=inv[mod%i]*1ll*(mod-mod/i)%mod;
  for(int i=2;i<500005;++i){
    inv[i]=inv[i-1]*1ll*inv[i]%mod;
  }
}
int C(int n,int m){
  return fac[n]*1ll*inv[m]%mod*inv[n-m]%mod;
}
int calc(int i,int j){
  int dlt1=xx[j]-xx[i],dlt2=yy[j]-yy[i];
  //double step1=,step2=;
  double step2=(dlt1*1.0*ay-dlt2*1.0*ax)/(bx*1.0*ay-by*1.0*ax);
  double step1=(dlt1*1.0*ay-step2*bx*ay)/ax/ay;
  //必然有解.无解状况已经排除.
  if(step1<0||step2<0)return 0;
  if(fabs(step1-floor(step1))>1e-8||fabs(step2-floor(step2))>1e-8)return 0;
  int n=step1,m=step2;
  return C(n+m,m);
}
int f[20][2];bool vis[20];
int g[20][20],deg[20];
int main(){
  freopen("knightmov.in","r",stdin);
  freopen("knightmov.out","w",stdout);
  int tests;scanf("%d",&tests);
  init();
  while(tests--){
    scanf("%d%d%d",&X,&Y,&K);
    scanf("%d%d%d%d",&ax,&ay,&bx,&by);
    for(int i=1;i<=K;++i)scanf("%d%d",xx+i,yy+i);
    if(ax==0&&ay==0&&bx==0&&by==0){
      if(X==0&Y==0){
	printf("-1\n");
      }else{
	printf("0\n");
      }
      continue;
    }
    dict.clear();
    for(int i=1;i<=K;++i)dict.insert(pair<int,int>(xx[i],yy[i]));
    if(ax==bx&&ay==by){
      printf("%d\n",check4(ax,ay));
    }
    else if(ax==0&&ay==0){
      printf("%d\n",check1(bx,by));
    }else if(bx==0&&by==0){
      printf("%d\n",check1(ax,ay));
    }else{
      double d1=direc(ax,ay),d2=direc(bx,by);
      if(d1==d2){
	printf("%d\n",check2());
      }else{
	if(X==0&&Y==0){
	  printf("1\n");continue;
	}else{
	  memset(f,0,sizeof(f));
	  xx[0]=0;yy[0]=0;xx[K+1]=X;yy[K+1]=Y;
	  f[0][0]=1;
	  for(int i=1;i<=K+1;++i)vis[i]=false;vis[0]=false;
	  memset(g,0,sizeof(g));memset(deg,0,sizeof(deg));
	  for(int i=0;i<=K+1;++i){
	    for(int j=0;j<=K+1;++j){
	      if(i!=j){
		g[i][j]=calc(i,j);
		if(g[i][j])deg[j]++;
	      }
	    }
	  }
	  for(int i=0;i<=K+1;++i){
	    int x=0x7f7f7f7f;
	    for(int j=0;j<=K+1;++j){
	      if(deg[j]==0&&!vis[j]){
		x=j;break;
	      }
	    }
	    if(x==0x7f7f7f7f)break;
	    vis[x]=true;
	    for(int t=0;t<=K+1;++t){
	      if(g[x][t]){
		deg[t]--;
		f[t][1]=(f[t][1]+f[x][0]*1ll*g[x][t]%mod)%mod;
		f[t][0]=(f[t][0]+f[x][1]*1ll*g[x][t]%mod)%mod;
	      }
	    }
	  }
	  printf("%d\n",(f[K+1][1]-f[K+1][0]+mod)%mod);
	}
      }
    }
  }
  // fclose(stdin);fclose(stdout);
  return 0;
}
/*
3
3 3 0
1 2 2 1
9 9 2
1 2 2 1
1 2 6 6
1 1 0
0 0 0 0

*/

posted @ 2017-06-10 16:21  liu_runda  阅读(384)  评论(0编辑  收藏  举报
偶然想到可以用这样的字体藏一点想说的话,可是并没有什么想说的. 现在有了:文化课好难