uva11732 "strcmp()" Anyone?

11732 “strcmp()” Anyone?

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:
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.
Input
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.
Output
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 ifUniversidad de Valladolid OJ: 11732 – “strcmp()” Anyone? 2/2
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.
Sample Input
2
a
b
4
cat
hat
mat
sir
0
Sample Output
Case 1: 1
Case 2: 6

题目大意:给出N个字符串(包含数字,字母),两两之间调用strcmp()相互比较,问字符比较的总次数.

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];}

思路:这里s[i] == t[i] 与 s[i] == '\0' 各有一次比较,所以相同的时候要比较两次,而如果s[i] != t[i]则只要比较一次.

那么就有ans += 现在与当前字符相同的字符串的个数*2 + 字符串总数- 现在与当前字符相同的字符串的个数(这个可以化简的我这里没有化简).

为了对应该公式,直接将Trie树的val值变成经过改点的字符串的个数就行,那么该节点需要比较的次数为2*val[u] + sum - val[u];

而且只要比较n*(n-1)/2次,那么每次插入一个字符串时,同时将它与前面的数比较就行.

相应代码:

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int sig_size = 64;const int maxnode = 4000*1000 + 100;long long ans;int sum;struct Trie{    int ch[maxnode][sig_size];    int val[maxnode];    int sz;    Trie(){        sz = 1;        memset(ch[0], 0, sizeof(ch));        memset(val, 0, sizeof(val));    }    int idx(char c){        if(c == '\0') return 26;        if(c <= '9' && c >= '0') return c-'0' + 27;        if(c <= 'z' && c >= 'a') return c-'a';        if(c <= 'Z' && c >= 'A') return c-'A'+37;    }    void clear(){        sz = 1;        memset(ch[0], 0, sizeof(ch));        memset(val, 0, sizeof(val));    }    void insert(char* s, int v = 1){        int u = 0, n = strlen(s);        int pre = sum-1;        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];            ans += val[u]*2;            ans += pre - val[u];            pre = val[u];            val[u] += v;        }    }    bool search(char* s, int len){        int u = 0;        for(int i = 0; i < len; i++){            int c = idx(s[i]);            if(!ch[u][c]) return false;            u = ch[u][c];        }        if(val[u]) return true;        else return false;    }};Trie T;int main(){    int n, cas = 1;    char s[1005];    while(~scanf("%d", &n)){        if(n == 0) break;        ans = sum = 0;        T.clear();        for(int i = 0; i < n; i++){            sum++;            scanf("%s", s);            T.insert(s);        }        printf("Case %d: %lld\n",cas++, ans);    }    return 0;}