strcmp() is a library function in C/C++ which compares two strings. It takes two strings as input parameter and decides which one is lexicographically larger or smaller: If the first string is greater then it returns a positive value, if the second string is greater it returns a negative value and if two strings are equal it returns a zero. The code that is used to compare two strings in C/C++ library is shown below:
int strcmp(char *s,char *t) { int i; for(i=0;s[i]==t[i];i++) if(s[i]=='\0') return 0; return s[i]-t[i]; }Figure: The standard strcmp() code provided for this problem. The number of comparisons required to compare two strings in strcmp() function is never returned by the function. But for this problem you will have to do just that at a larger scale. strcmp() function continues to compare characters in the same position of the two strings until two different characters are found or both strings come to an end. Of course it assumes that last character of a string is a null (‘\0’) character. For example the table below shows what happens when “than” and “that”; “therE” and “the” are compared using strcmp() function. To understand how 7 comparisons are needed in both cases please consult the code block given above.
The input file contains maximum 10 sets of inputs. The description of each set is given below: Each set starts with an integer N (0 < N < 4001) which denotes the total number of strings. Each of the next N lines contains one string. Strings contain only alphanumerals (‘0’...‘9’, ‘A’...‘Z’, ‘a’...‘z’) have a maximum length of 1000, and a minimum length of 1. Input is terminated by a line containing a single zero.
For each set of input produce one line of output. This line contains the serial of output followed by an integer T. This T denotes the total number of comparisons that are required in the strcmp() function if all the strings are compared with one another exactly once. So for N strings the function strcmp() will be called exactly N(N−1)/2 times. You have to calculate total number of comparisons inside the strcmp() function in those N(N−1)/2 calls. You can assume that the value of T will fit safely in a 64-bit signed integer. Please note that the most straightforward solution (Worst Case Complexity O(N²*1000) will time out for this problem.
2 a b 4 cat hat mat sir 0
Case 1: 1 Case 2: 6
这题一开始看还是挺简单的。 一眼看出用trie，到树的每一个分叉计算子树互相比较次数 但是细节好多啊！！！
要分3种情况：
设单词长度为len 则比较次数为2×(len+1)
设短的单词长度为len 则比较次数为2×len+1
设两个单词相同部分长度为len 则比较次数为len×2+1
trie每个节点要记下flag（有多少单词在当前点结束）与size（当前点及其子树中共多少单词） 每种情况在计算中都要判断
代码： （注意：答案要转long long）
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> using namespace std; const int N = 4005; typedef long long ll; struct trie { trie *ch[65]; int flag,size; void clear(){ flag=size=0; for(int i=0;i<63;i++) ch[i]=NULL; } }pool[N*1005],*root; int cnt; int get(char s){ if(s>='0' && s<='9') return s-'0'; else if(s>='A' && s<='Z') return s-'A'+10; return s-'a'+36; } void add(){ char s[1005]; scanf("%s",s); int len=strlen(s),id; trie *p=root; p->size++; for(int i=0;i<len;i++){ id=get(s[i]); if(!p->ch[id]){ pool[++cnt].clear(); p->ch[id]=&pool[cnt]; } p=p->ch[id]; p->size++; } p->flag++; } ll dfs(trie *p,int dep){ int size=0; ll ret=0; size=p->size-p->flag; for(int i=0;i<63;i++) if(p->ch[i]) ret+=((ll)size-p->ch[i]->size)*p->ch[i]->size*((ll)dep*2+1)+2*dfs(p->ch[i],dep+1); ret/=2; if(p->flag){ ret+=(ll)p->flag*(p->flag-1)*(dep+1); ret+=(ll)p->flag*size*(dep*2+1); } return ret; } int n; int main() { int i,kase=0; root=&pool[++cnt]; while(scanf("%d",&n) && n){ cnt=1; pool[1].clear(); for(i=0;i<n;i++) add(); printf("Case %d: %lld\n",++kase,dfs(root,0)); } return 0; }转载于:https://www.cnblogs.com/lindalee/p/8179283.html
