题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3065
3 AA BB CC ooxxCC%dAAAoen....END
AA: 2 CC: 1HintHit: 题目描述中没有被提及的所有情况都应该进行考虑。比如两个病毒特征码可能有相互包含或者有重叠的特征码段。 计数策略也可一定程度上从Sample中推测。
代码如下:
#include <stdio.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> #include <string> using namespace std; char ss[1017][57]; int ans[1017]; struct Trie { int next[500010][128],fail[500010],end[500010]; int root,L; int newnode() { for(int i = 0; i < 128; i++) next[L][i] = -1; end[L++] = -1; return L-1; } void init() { L = 0; root = newnode(); } void insert(char buf[], int id) { int len = strlen(buf); int now = root; for(int i = 0; i < len; i++) { if(next[now][buf[i]] == -1) next[now][buf[i]] = newnode(); now = next[now][buf[i]]; } end[now] = id; } void build() { queue<int>Q; fail[root] = root; for(int i = 0; i < 128; i++) if(next[root][i] == -1) next[root][i] = root; else { fail[next[root][i]] = root; Q.push(next[root][i]); } while(!Q.empty()) { int now = Q.front(); Q.pop(); for(int i = 0; i < 128; i++) if(next[now][i] == -1) next[now][i] = next[fail[now]][i]; else { fail[next[now][i]]=next[fail[now]][i]; Q.push(next[now][i]); } } } void query(char buf[], int n) { memset(ans,0,sizeof(ans)); int len = strlen(buf); int now = root; for(int i = 0; i < len; i++) { now = next[now][buf[i]]; int temp = now; while( temp != root ) { //res += end[temp]; //printf("temp:%d end[temp]:%d\n",temp,end[temp]); if(end[temp] != -1) { ans[end[temp]]++; } //end[temp] = 0; temp = fail[temp]; } } } }; char buf[2000010]; Trie ac; int main() { int t, n; while(~scanf("%d",&n)) { ac.init(); for(int i = 0; i < n; i++) { scanf("%s",ss[i]); ac.insert(ss[i],i); } ac.build();//fail scanf("%s",buf); ac.query(buf, n); for(int i = 0; i < n; i++) { if(ans[i]) { printf("%s: %d\n",ss[i],ans[i]); } } } return 0; }
原文地址:http://blog.csdn.net/u012860063/article/details/44465401