题目来源:Light OJ 1114 Easily Readable
题意:求一个句子有多少种组成方案 只要满足每个单词的首尾字符一样 中间顺序可以变化
思路:每个单词除了首尾 中间的字符排序 然后插入字典树 记录每个单词的数量
输入一个句子 每个单词也排序之后查找 根据乘法原理 答案就是每个单词的数量之积
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; const int maxnode = 111111; const int sigma_size = 52; int ch[maxnode][sigma_size]; int val[maxnode]; int sz; void init() { sz = 1; memset(ch[0], -1, sizeof(ch[0])); } int idx(char c) { if(c >= 'a' && c <= 'z') return c - 'a'; else return c - 'A' + 26; } 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] == -1) { memset(ch[sz], -1, sizeof(ch[sz])); val[sz] = 0; ch[u][c] = sz++; } u = ch[u][c]; } val[u]++; } int find(char *s) { int u = 0, n = strlen(s); for(int i = 0; i < n; i++) { int c = idx(s[i]); u = ch[u][c]; if(val[u] == -1) return 0; } return val[u]; } int main() { int cas = 1; int T; scanf("%d", &T); while(T--) { init(); int n; scanf("%d", &n); for(int i = 0; i < n; i++) { char s[11111]; scanf("%s", s); int len = strlen(s); if(len > 2) sort(s+1, s+len-1); insert(s); } scanf("%d", &n); getchar(); printf("Case %d:\n", cas++); while(n--) { int sum = 1; char s[11111]; gets(s); char *p = strtok(s, " "); while(p) { char str[11111]; strcpy(str, p); int len = strlen(str); if(len > 2) sort(str+1, str+len-1); sum *= find(str); p = strtok(NULL, " "); } printf("%d\n", sum); } } return 0; }
Light OJ 1114 Easily Readable 字典树,布布扣,bubuko.com
Light OJ 1114 Easily Readable 字典树
原文地址:http://blog.csdn.net/u011686226/article/details/28098577