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UVa 11468 Substring (AC自动机+概率DP)
题意:给出一个字母表以及每个字母出现的概率。再给出一些模板串S。从字母表中每次随机拿出一个字母,一共拿L次组成一个产度为L的串,
问这个串不包含S中任何一个串的概率为多少?
析:先构造一个AC自动机,然后随机生成L个字母,就是在AC自动机的某个结点走多少步,dp[i][j] 表示在 i 结点,并且剩下 j 步,
然后记忆化搜索就OK了。
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> #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 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 = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const int mod = 1e9 + 7; 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 = 20 * 20 + 10; const int sigma_size = 128; struct Aho_Corasick{ int ch[maxnode][sigma_size]; int f[maxnode]; bool match[maxnode]; int sz; void init(){ sz = 1; memset(ch[0], 0, sizeof ch[0]); } void insert(char *s){ int u = 0; for(int i = 0; s[i]; ++i){ int c = s[i]; if(!ch[u][c]){ memset(ch[sz], 0, sizeof ch[sz]); match[sz] = 0; ch[u][c] = sz++; } u = ch[u][c]; } match[u] = true; } int getFail(){ queue<int> q; f[0] = 0; for(int c = 0; c < sigma_size; ++c){ int u = ch[0][c]; if(u){ f[u] = 0; q.push(u); } } while(!q.empty()){ int r = q.front(); q.pop(); for(int c = 0; c < sigma_size; ++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]; match[u] |= match[f[u]]; } } } }; Aho_Corasick ac; double dp[maxnode][110]; bool vis[maxnode][110]; char s[100]; struct Node{ int x; double p; }; vector<Node> v; double dfs(int u, int L){ if(!L) return 1.0; double &ans = dp[u][L]; if(vis[u][L]) return ans; vis[u][L] = 1; ans = 0.0; for(int i = 0; i < n; ++i){ int c = v[i].x; if(!ac.match[ac.ch[u][c]]) ans += dfs(ac.ch[u][c], L-1) * v[i].p; } return ans; } int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d", &n); ac.init(); for(int i = 0; i < n; ++i){ scanf("%s", s); ac.insert(s); } ac.getFail(); scanf("%d", &n); v.clear(); for(int i = 0; i < n; ++i){ double pp; scanf("%s %lf", s, &pp); v.push_back((Node){s[0], pp}); } int L; scanf("%d", &L); memset(vis, 0, sizeof vis); printf("Case #%d: %f\n", kase, dfs(0, L)); } return 0; }
UVa 11468 Substring (AC自动机+概率DP)
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