HDU 5384 Danganronpa (AC自己主动机模板题)
题意:给出n个文本和m个模板。求每一个文本中全部模板出现的总次数。
思路:Trie树权值记录每一个模板的个数。对于每一个文本跑一边find就可以。
#include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<map> #include<ctime> #include<set> #define eps 1e-6 #define LL long long #define pii (pair<int, int>) //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; //const int INF = 0x3f3f3f3f; int n, k, shit; const int maxn = 1000000 + 100; const int SIGMA_SIZE = 26; const int maxnode = 1000000+100; int ch[maxnode][SIGMA_SIZE+5]; int val[maxnode], cnt[100000+1000]; int idx(char c) {return c - 'a';} struct Trie { int sz; Trie() { sz = 1; memset(ch[0], 0, sizeof(ch[0]));}; void insert(char *s) { int u = 0, n = strlen(s); for(int i = 0; i < n; i++) { int c = idx(s[i]); if(!ch[u][c]) { memset(ch[sz], 0, sizeof(ch[sz])); val[sz] = 0; ch[u][c] = sz++; } u = ch[u][c]; } val[u]++; } }; //ac自己主动机 int last[maxn], f[maxn]; void print(int i, int j) { if(j) { shit += val[j]; //cout << i << endl; print(i, last[j]); } } 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); last[u] = 0; } } 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]; last[u] = val[f[u]] ? f[u] : last[f[u]]; } } } void find_T(char* T) { int len = strlen(T); //cout << len << endl; int j = 0; for(int i = 0; i < len; i++) { int c = idx(T[i]); j = ch[j][c]; if(val[j]) print(i, j); else if(last[j]) print(i, last[j]); } } string text[100000+100]; char tmp[10050], tt[105000]; int ans[100000+10000]; int main() { //freopen("input.txt", "r", stdin); int T; cin >> T; while(T--) { scanf("%d%d", &n, &k); memset(ans, 0, sizeof(ans[0])*(n+10)); Trie trie; for(int i = 0; i < n; i++) { scanf("%s", tmp); text[i] = tmp; } for(int i = 0; i < k; i++) { scanf("%s", tmp); //cout << tmp << endl; trie.insert(tmp); } getFail(); for(int i = 0; i < n; i++) { shit = 0; find_T((char*)text[i].c_str()); ans[i] = shit; } for(int i = 0; i < n; i++) printf("%d\n", ans[i]); } return 0; }
posted on 2017-06-06 18:03 yjbjingcha 阅读(115) 评论(0) 编辑 收藏 举报