Codeforces Round #316 DIV2 E問題 - Pig and Palindromes

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Codeforces Round #316 DIV2 E問題 (2500pt)
Problem description

問題概要

$N \times M$ の盤面が与えられる．
各マスにはアルファベット小文字が $1$ 文字書かれている．
左上のマスから右下のマスへと移動する経路の中で，右か下のどちらかにしか進まないもので，通ったマスに書かれているアルファベットを繋げると回文になるもののパターンの数を ${\rm mod}\ 1000000007\ (10^9+7)$ で求める問題．

解法

左上と右下から同時にスタートして，同じ文字のマスしか移動できないとして，最終的に合流するパターンの数を数えれば良い．
動的計画法で，それぞれの場所のペアを状態として，パターン数を計算していけば良い．
メモリは，移動回数ごとに配列を使いまわすようにすれば，移動回数は $500$ 回程度，同じ移動回数での状態数は $500^2$ 程度なので，
時間計算量は $O(NM^2)$，空間計算量は $O(M^2)$ 程度になる．

C++のスパゲッティなコード

#include<bits/stdc++.h>
using namespace std;

#define REP(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

#define mygc(c) (c)=getchar()
#define mypc(c) putchar(c)

#define ll long long
#define ull unsigned ll

void reader(int *x){int k,m=0;*x=0;for(;;){mygc(k);if(k=='-'){m=1;break;}if('0'<=k&&k<='9'){*x=k-'0';break;}}for(;;){mygc(k);if(k<'0'||k>'9')break;*x=(*x)*10+k-'0';}if(m)(*x)=-(*x);}
int reader(char c[]){int i,s=0;for(;;){mygc(i);if(i!=' '&&i!='\n'&&i!='\r'&&i!='\t'&&i!=EOF) break;}c[s++]=i;for(;;){mygc(i);if(i==' '||i=='\n'||i=='\r'||i=='\t'||i==EOF) break;c[s++]=i;}c[s]='\0';return s;}
template <class T, class S> void reader(T *x, S *y){reader(x);reader(y);}

void writer(int x, char c){int s=0,m=0;char f[10];if(x<0)m=1,x=-x;while(x)f[s++]=x%10,x/=10;if(!s)f[s++]=0;if(m)mypc('-');while(s--)mypc(f[s]+'0');mypc(c);}
template<class T> void writerLn(T x){writer(x,'\n');}

char memarr[47000000]; void *mem = memarr;
#define MD 1000000007

int get_inv(ll a, int md){ll t=a,s=md,u=1,v=0,e;while(s){e=t/s;t-=e*s;u-=e*v;swap(t,s);swap(u,v);}if(u<0)u+=md;return u;}

struct mint{
  static unsigned md, W, R, Rinv, mdninv, RR;
  unsigned val;

  mint(){}mint(int a){val=mulR(a);}mint(unsigned a){val=mulR(a);}mint(ll a){val=mulR(a);}mint(ull a){val=mulR(a);}
  unsigned setmod(unsigned m){int i;unsigned t;W=32;md=m;R=(1ULL<<W)%md;RR=(ull)R*R%md;switch(m){case 104857601:Rinv=2560000;mdninv=104857599;break;case 998244353:Rinv=232013824;mdninv=998244351;break;case 1000000007:Rinv=518424770;mdninv=2226617417U;break;case 1000000009:Rinv=171601999;mdninv=737024967;break;case 1004535809:Rinv=234947584;mdninv=1004535807;break;case 1007681537:Rinv=236421376;mdninv=1007681535;break;case 1012924417:Rinv=238887936;mdninv=1012924415;break;case 1045430273:Rinv=254466304;mdninv=1045430271;break;case 1051721729:Rinv=257538304;mdninv=1051721727;break;default:Rinv=get_inv(R,md);mdninv=0;t=0;rep(i,W){if(t%2==0)t+=md,mdninv|=(1U<<i);t/=2;}}}
  unsigned mulR(unsigned a){return(ull)a*R%md;}unsigned mulR(int a){if(a<0)a=a%md+md;return mulR((unsigned)a);}unsigned mulR(ull a){return mulR((unsigned)(a%md));}unsigned mulR(ll a){if(a<0)a=a%md+md;return mulR((unsigned)a);}
  unsigned reduce(unsigned T){unsigned m=T*mdninv;unsigned t=(unsigned)((T+(ull)m*md)>>W);if(t>=md)t-=md;return t;}unsigned reduce(ull T){unsigned m=(unsigned)T*mdninv;unsigned t=(unsigned)((T+(ull)m*md)>>W);if(t>=md)t-=md;return t;}
  unsigned get(){return reduce(val);}
  mint&operator+=(mint a){val+=a.val;if(val>=md)val-=md;return*this;}mint&operator-=(mint a){if(val<a.val)val=val+md-a.val;else val-=a.val;return*this;}mint&operator*=(mint a){val=reduce((ull)val*a.val);return*this;}mint&operator/=(mint a){return*this*=a.inverse();}
  mint operator+(mint a){return mint(*this)+=a;}mint operator-(mint a){return mint(*this)-=a;}mint operator*(mint a){return mint(*this)*=a;}mint operator/(mint a){return mint(*this)/=a;}
  mint operator+(int a){return mint(*this)+=mint(a);}mint operator-(int a){return mint(*this)-=mint(a);}mint operator*(int a){return mint(*this)*=mint(a);}mint operator/(int a){return mint(*this)/=mint(a);}
  mint operator+(ll a){return mint(*this)+=mint(a);}mint operator-(ll a){return mint(*this)-=mint(a);}mint operator*(ll a){return mint(*this)*=mint(a);}mint operator/(ll a){return mint(*this)/=mint(a);}
  mint operator-(void){mint res;if(val)res.val=md-val;else res.val=0;return res;}
  operator bool(void){return val!=0;}operator int(void){return get();}operator ll(void){return get();}
  mint inverse(){int a=val,b=md,u=1,v=0,t;mint r;while(b){t=a/b;a-=t*b;swap(a,b);u-=t*v;swap(u,v);}if(u<0)u+=md;r.val=(ull)u*RR%md;return r;}
  mint pw(ull b){mint a(*this),r;r.val=R;while(b){if(b&1)r*=a;b>>=1;a*=a;}return r;}
  bool operator==(int a){return get()==a;}
  bool operator!=(int a){return get()!=a;}
};
unsigned mint::md, mint::W, mint::R, mint::Rinv, mint::mdninv, mint::RR;
mint operator+(int a, mint b){return mint(a)+=b;}mint operator-(int a, mint b){return mint(a)-=b;}mint operator*(int a, mint b){return mint(a)*=b;}mint operator/(int a, mint b){return mint(a)/=b;}
mint operator+(ll a, mint b){return mint(a)+=b;}mint operator-(ll a, mint b){return mint(a)-=b;}mint operator*(ll a, mint b){return mint(a)*=b;}mint operator/(ll a, mint b){return mint(a)/=b;}

mint mval[10000], minv[10000];
void mint_init(int md=MD, mint val[]=mval, int vals=10000, mint inv[]=minv, int invs=10000){int i;val[0].setmod(md);val[0].val=0;REP(i,1,vals){val[i].val=val[i-1].val+mint::R;if(val[i].val >=md)val[i].val-=md;}inv[1].val=1;REP(i,2,invs){inv[i].val=md-((ll)(md/i)*inv[md%i].val%md);}REP(i,1,invs)inv[i].val=(ull)inv[i].val*mint::R%md;}

template<class T> void *malloc2d(T ***arr, int x, int y, void *mem){int i;(*arr)=(T**)mem;(*arr)[0]=(T*)((*arr)+x);REP(i,1,x)(*arr)[i]=(*arr)[i-1]+y;return((*arr)[x-1]+y);}

int X, Y;
char mp[555][555];

mint **dp, **nx;

int main(){
  int i, j, k, a, b, d;
  int x1, y1, x2, y2;
  int x11, y11, x22, y22;
  int step;
  mint res;

  mint_init();

  reader(&X,&Y);
  rep(i,X) reader(mp[i]);
  mem = malloc2d(&dp, X, X, mem);
  mem = malloc2d(&nx, X, X, mem);

  rep(i,X) rep(j,X) dp[i][j] = mval[0];
  rep(i,X) rep(j,X) nx[i][j] = mval[0];
  if(mp[0][0]==mp[X-1][Y-1]) dp[0][X-1] = mval[1];

  step = X+Y-2;
  for(k=0;k+1<step;k+=2){
    rep(x1,X) REP(x2,x1,X) if(dp[x1][x2].val){
      y1 = k/2 - x1;
      y2 = Y-1 - (k/2 - (X-1 - x2));
      rep(d,4){
        x11 = x1; x22 = x2;
        y11 = y1; y22 = y2;
        if(d<2) x11++; else y11++;
        if(d%2) x22--; else y22--;
        if(x11 >= X || x22 < 0) continue;
        if(y11 >= Y || y22 < 0) continue;
        if(mp[x11][y11] != mp[x22][y22]) continue;
        nx[x11][x22] += dp[x1][x2];
      }
      dp[x1][x2].val = 0;
    }

    swap(dp, nx);
  }

  res = mval[0];
  rep(i,X) res += dp[i][i];
  if(k==step-1){
    REP(i,1,X) res += dp[i-1][i];
  }

  writerLn( (int)res );

  return 0;
}

Current time: 2024年04月20日16時45分44秒
Last modified: 2015年08月14日23時34分33秒 (by laycrs)
Tags: Competitive_Programming Codeforces CF316 CF_Div2_E
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