http://acm.hdu.edu.cn/showproblem.php?pid=2222
就是裸的多串匹配的问题,AC自动机模板题
http://www.cppblog.com/MatoNo1/archive/2011/10/19/158635.html
用了yang_7_46的模板
#include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue> #include <map> #include <iostream> #include <algorithm> using namespace std; #define RD(x) scanf("%d",&x) #define RD2(x,y) scanf("%d%d",&x,&y) #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define clr0(x) memset(x,0,sizeof(x)) #define eps 1e-9 const double pi = acos(-1.0); typedef long long LL; const int modo = 1e9 + 7; const int maxn = 1e6 + 5,maxm = 1e4 + 5; int cnt; struct AC_Automata { #define N 240005 #define M 26 int ch[N][M], sz; int val[N], last[N], f[N], cc[N]; void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); } int idx(char c) { return c-'a'; } void insert(char s[], int v) { int u = 0; for (int i=0; s[i]; i++) { int c = idx(s[i]); if (!ch[u][c]) { memset(ch[sz], 0, sizeof(ch[sz])); val[sz] = cc[sz] = 0; ch[u][c] = sz++; } u = ch[u][c]; } cc[u]++; val[u] = v; } void build() { queue<int> q; f[0] = 0; for (int c=0; c<M; c++) { int u = ch[0][c]; if (u) { f[u] = 0, q.push(u); last[u] = 0; } } while (!q.empty()) { int r = q.front(); q.pop(); for (int c=0; c<M; c++) { int u = ch[r][c]; if (!u) { ch[r][c] = ch[f[r]][c]; continue; } q.push(u); int v = f[r]; f[u] = ch[v][c]; last[u] = val[f[u]] ? f[u] : last[f[u]]; } } } void find(char *s) { int j = 0; for (int i=0; s[i]; i++) { int c = idx(s[i]); j = ch[j][c]; //if (val[j]) print(j); //else if (last[j]) print(last[j]); if (cc[j]) print(j); else if (last[j]) print(last[j]); } } void print(int j) { //打印以结点j结尾的所有字符串 if (j) { //printf("%d: %d ", j, val[j]); cnt += cc[j]; cc[j] = 0; print(last[j]); } } } ac; char s[55], text[1000006]; int main() { int n, _; RD(_); while (_--) { RD(n); ac.clear(); for (int i=1; i<=n; i++) { scanf(" %s", s); ac.insert(s, i); } ac.build(); cnt = 0; scanf(" %s", text); ac.find(text); printf("%d ", cnt); } return 0; }