摘要:
"POJ" 哈夫曼树模板题,记得开long long就行了. cpp // include include include include include include include using namespace std; inline int read(){ int x=0,o=1;char 阅读全文
摘要:
"POJ" 题面懒得放了,就是一道 "哈夫曼树" 的模板题,而且还是2叉的,直接丢代码了. cpp // include include include include include include include using namespace std; inline int read(){ i 阅读全文
摘要:
"AcWing" 题意:给定一张N个点M条边的有向无环图,分别统计从每个点出发能够到达的点的数量.$(N,M include include include include include include include using namespace std; inline int read(){ 阅读全文
摘要:
分析:对于第一问判断只需要倍长其中一个字符串(相当于要断环为链),然后哈希两个字符串,最后$O(N)$扫描是否有一段长度为n的字符串的哈希值相同即可.对于第二问需要用到“字符串的最小表示法”,临时在 "博客" 上学了一下,这个算法也是$O(N)$的. cpp // include include i 阅读全文
摘要:
"POJ" 题意:有一个商店有许多批货,每一批货又有N(0 include include include include include include using namespace std; inline int read(){ int x=0,o=1;char ch=getchar(); w 阅读全文