[AC自动机 概率DP 矩阵乘法||高斯消元] BZOJ 1444 [Jsoi2009]有趣的游戏

AC自动机建转移矩阵 

然后

要么矩阵乘法 无限迭代 数据范围小可以接受

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long double ld;
//typedef double ld;

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 int read(char *s)
{
char c=nc(); int len=0;
for (;!(c>='A' && c<='Z');c=nc());
for (;c>='A' && c<='Z';s[++len]=c,c=nc()); s[++len]=0; return len-1;
}

const int N=15;
const int M=105;

int root,ncnt,ch[M][26];
int pos
,f[M],val[M];

inline void Insert(int len,char *str,int idx){
int p=root;
for (int i=1;i<=len;i++)
{
if (!ch[p][str[i]-'A']) ch[p][str[i]-'A']=++ncnt;
p=ch[p][str[i]-'A'];
}
pos[idx]=p; val[p]=1;
}

int Q[M],l,r;

inline void GetFail(){
f[0]=0; l=r=-1; int u,v,t;
for (int i=0;i<26;i++)
if (ch[0][i])
f[ch[0][i]]=0,Q[++r]=ch[0][i];
while (l<r){
u=Q[++l];
for (int i=0;i<26;i++)
{
v=ch[u][i];
if (!v) { ch[u][i]=ch[f[u]][i]; continue; }
Q[++r]=v;
f[v]=ch[f[u]][i];
}
}
}

struct Matrix{
ld a[M][M]; int n;
Matrix(int _n=0,int p=0){
n=_n;
for (int i=0;i<=n;i++)
for (int j=0;j<=n;j++)
a[i][j]=0;
if (p) for (int i=0;i<=n;i++) a[i][i]=1;
}
ld *operator [](int x){
return a[x];
}
friend Matrix operator * (Matrix &A,Matrix &B){
int n=A.n; Matrix ret(n);
for (int i=0;i<=n;i++)
for (int j=0;j<=n;j++)
for (int k=0;k<=n;k++)
ret[i][j]+=A[i][k]*B[k][j];
return ret;
}
friend Matrix Pow(Matrix a,int b){
Matrix ret(a.n,1);
for (;b;b>>=1,a=a*a)
if (b&1)
ret=ret*a;
return ret;
}
}A;

int len,n,m;
ld p[26];

int main()
{
int ip,iq; char str[15];
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); read(len); read(m);
for (int i=0;i<m;i++)
read(ip),read(iq),p[i]=(ld)ip/iq;
for (int i=1;i<=n;i++)
read(str),Insert(len,str,i);
GetFail();
A=Matrix(ncnt);
for (int i=0;i<=ncnt;i++)
{
if (val[i]){
A[i][i]=1;
continue;
}
for (int j=0;j<26;j++) A[i][ch[i][j]]+=p[j];
}
for (int i=1;i<=50;i++)
A=A*A;
for (int i=1;i<=n;i++)
printf("%.2lf\n",(double)A[0][pos[i]]);
return 0;
}

要么高斯消元求精确解

http://y4612s.logdown.com/posts/192615-solution-bzoj1444-jsoi2009-fun-game

http://www.cnblogs.com/zig-zag/archive/2013/05/15/3080764.html

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#define dprintf(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long double ld;
//typedef double ld;

const ld eps=1e-10;

inline int dcmp(ld a,ld b){
if (fabs(a-b)<eps) return 0;
if (a<b) return -1; return 1;
}

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 int read(char *s)
{
char c=nc(); int len=0;
for (;!(c>='A' && c<='Z');c=nc());
for (;c>='A' && c<='Z';s[++len]=c,c=nc()); s[++len]=0; return len-1;
}

const int N=15;
const int M=105;

int root,ncnt,ch[M][26];
int pos
,f[M],val[M];

inline void Insert(int len,char *str,int idx){
int p=root;
for (int i=1;i<=len;i++)
{
if (!ch[p][str[i]-'A']) ch[p][str[i]-'A']=++ncnt;
p=ch[p][str[i]-'A'];
}
pos[idx]=p; val[p]=1;
}

int Q[M],l,r;

inline void GetFail(){
f[0]=0; l=r=-1; int u,v,t;
for (int i=0;i<26;i++)
if (ch[0][i])
f[ch[0][i]]=0,Q[++r]=ch[0][i];
while (l<r){
u=Q[++l];
for (int i=0;i<26;i++)
{
v=ch[u][i];
if (!v) { ch[u][i]=ch[f[u]][i]; continue; }
Q[++r]=v;
f[v]=ch[f[u]][i];
}
}
}

int len,n,m;
ld p[26];

ld A[M][M],ans[M];

inline void Print()
{
for (int i=1;i<=m;i++,dprintf("\n"))
for (int j=1;j<=n+1;j++)
dprintf("%.3lf ",(double)(A[i][j]+eps));
dprintf("\n");
}

inline void Gauss()
{
int l=1,r; double tmp;
// Print();
for (int i=1;i<=m;i++)
{
for (r=l;r<=m;r++) if (dcmp(A[r][i],0.0)!=0) break;
if (r==m+1) continue;
if (l!=r) for (int j=1;j<=n+1;j++) swap(A[l][j],A[r][j]);
tmp=A[l][i];
for (int j=1;j<=n+1;j++)
A[l][j]/=tmp;
for (int j=1;j<=m+1;j++)
{
if (j==l) continue;
tmp=A[j][i];
if (tmp)
for (int k=1;k<=n+1;k++)
A[j][k]-=tmp*A[l][k];
}
l++;
// Print();
}
for (int i=1;i<=n;i++)
ans[i-1]=A[i][n+1];
}

int main()
{
int ip,iq; char str[15]; int in;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); read(len); read(m);
for (int i=0;i<m;i++)
read(ip),read(iq),p[i]=(ld)ip/iq;
int flag=1;
for (int i=1;i<=n;i++)
{
read(str),Insert(len,str,i);
int f1=0;
for (int j=1;j<=len;j++)
if (dcmp(p[str[j]-'A'],0)==0)
f1=1;
flag&=f1;
}
if (flag) { for (int i=1;i<=n;i++) printf("0.00\n"); return 0; }
GetFail();
for (int i=0;i<=ncnt;i++)
{
if (val[i]) continue;
for (int j=0;j<26;j++)
A[ch[i][j]+1][i+1]+=p[j];
}
for (int i=0;i<=ncnt;i++)
A[i+1][i+1]-=1;
for (int i=0;i<=ncnt;i++)
A[1][i+1]=val[i];
A[1][ncnt+2]=1;
in=n; n=ncnt+1; m=ncnt+1;
Gauss();
for (int i=1;i<=in;i++)
printf("%.2lf\n",(double)(ans[pos[i]]+eps));
return 0;
}