1444: [Jsoi2009]有趣的游戏
Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 1382 Solved: 498
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Description
Input
注意 是0<=P
Output
Sample Input
Sample Output
HINT
30%的数据保证, n ≤ 2. 50%的数据保证, n ≤ 5. 100%的数据保证, n , l, m≤ 10.
Source
析：很容易列出方程，dp[i] = ∑dp[j] * pj ，所以要处理出来就需要AC自动机，然后再用Gauss 消元即可。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 150000 + 10;
const int maxm = 3e5 + 10;
const int mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
const int maxnode = 10 * 10 + 50;
int sigma;
double A[maxnode][maxnode];
double p[15];
int pos[15];
struct Aho{
int ch[maxnode][11], f[maxnode];
bool val[maxnode];
int sz;
void init(){ sz = 1; ms(ch[0], 0); }
inline int idx(char ch){ return ch - 'A'; }
int insert(const char *s){
int u = 0;
for(int i = 0; s[i]; ++i){
int c = idx(s[i]);
if(!ch[u][c]){
ms(ch[sz], 0);
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
val[u] = 1;
return u;
}
void getFail(){
queue<int> q; f[0] = 0;
for(int c = 0; c < sigma; ++c){
int u = ch[0][c];
if(u){ q.push(u); f[u] = 0; }
}
while(!q.empty()){
int r = q.front(); q.pop();
for(int c = 0; c < sigma; ++c){
int u = ch[r][c];
if(!u){ ch[r][c] = ch[f[r]][c]; continue; }
q.push(u);
int v = f[r];
while(v && !ch[v][c]) v = f[v];
f[u] = ch[v][c];
}
}
}
int solve(){
for(int i = 0; i < sz; ++i){
A[i][i] += 1.;
if(val[i]) continue;
for(int j = 0; j < sigma; ++j){
int nxt = ch[i][j];
A[nxt][i] -= p[j];
}
}
return sz;
}
};
Aho aho;
char s[20];
void Gauss(int n){
for(int i = 0; i < n; ++i){
int r = i;
for(int j = i+1; j < n; ++j)
if(fabs(A[j][i] > fabs(A[r][i]))) r = j;
if(r != i) for(int j = 0; j <= n; ++j) swap(A[r][j], A[i][j]);
for(int k = i+1; k < n; ++k){
double f = A[k][i] / A[i][i];
for(int j = i; j <= n; ++j) A[k][j] -= f * A[i][j];
}
}
for(int i = n-1; i >= 0; --i){
for(int j = i+1; j < n; ++j)
A[i][n] -= A[j][n] * A[i][j];
A[i][n] /= A[i][i];
}
}
int main(){
scanf("%d %d %d", &n, &m, &sigma);
for(int i = 0; i < sigma; ++i){
int x, y; scanf("%d %d", &x, &y);
p[i] = x * 1. / y;
}
aho.init();
for(int i = 1; i <= n; ++i){
scanf("%s", s);
pos[i] = aho.insert(s);
}
aho.getFail();
int len = aho.solve();
A[0][len] = 1.;
Gauss(len);
for(int i = 1; i <= n; ++i) printf("%.2f
", A[pos[i]][len] / A[pos[i]][pos[i]]);
return 0;
}