UVa 11732 - strcmp() Anyone?

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2832

strcmp() Anyone? Time Limit:2000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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Appoint description:

Description

J	“strcmp()” Anyone? Input: Standard Input Output: Standard Output

 

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.

 

t

h

a

N

\0

t

h

e

r

E

\0

=

=

=

≠

=

=

=

≠

t

h

a

T

\0

t

h

e

\0

Returns negative value 7 Comparisons

Returns positive value 7 Comparisons

 

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. Input file size is around 23 MB.

 

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 if all the strings are compared with one another exactly once. So for N strings the function strcmp() will be called exactly  times. You have to calculate total number of comparisons inside the strcmp() function in those  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                              Output for Sample Input

2 a b 4 cat hat mat sir 0

Case 1: 1 Case 2: 6

题目大意：输入n个字符串（ 1~9,a~z,A~Z组成）.每两个字符串比较，输出总的比较次数。

由于结点数量较大，普通的建树方法会超时（初始化）。这里用左孩子右兄弟的建树方法实现trie

思路：用数组记录当前字母出现的次数t[v]，每次比较的时候 如果相同ans+=2*t[v];

    p[v]记录v后面的'\0'数量。具体见代码；

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 4001000
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b)
{
    return a>b;
}

struct Trie
{
	char ch[maxn];
	int left[maxn],right[maxn],cnt;
	int t[maxn];
	int p[maxn];
	ll ans;
	ll k;
	Trie(){}
	void init()
	{
		ans=0;
		cnt=1;
		left[0]=right[0]=0;
		t[0]=0;
		k=0;
	}
	void add(const char*s)
	{
		ans+=k;
		k++;
		int len=strlen(s),u=0,v;
		for(int i=0;i<len;i++)
		{
			for(v=left[u];v;v=right[v])
			if(ch[v]==s[i])
				break;
			if(v==0)
			{
				v=cnt++;
				p[v]=t[v]=0;
				ch[v]=s[i];
				right[v]=left[u];
				left[u]=v;
				left[v]=0;
			}
			u=v;
			ans+=2*t[v];
			t[v]++;
		}
		ans+=p[u];
		p[u]++;
	}
}tree;

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    char s[2000];
    int n;
    int cnt=1;
    while(scanf("%d",&n)&&n)
	{
		tree.init();
		for(int i=1;i<=n;i++)
		{
			scanf("%s",s);
			tree.add(s);
		}
		 printf("Case %d: %lld\n",cnt++,tree.ans);
	}
    return 0;
}