题目描写：
Autocomplete25 points
Since you crave state-of-the-art technology, you've just purchased a phone with a great new feature: autocomplete! Your phone's version of autocomplete has some pros and cons. On the one hand, it's very cautious.
It only autocompletes a word when it knows exactly what you're trying to write. On the other hand, you have to teach it every word you want to use.
You have N distinct words that you'd like to send in a text message in order. Before sending each word, you add it to your phone's dictionary. Then, you write the smallest non-empty prefix of the
word necessary for your phone to autocomplete the word. This prefix must either be the whole word, or a prefix which is not a prefix of any other word yet in the dictionary.
What's the minimum number of letters you must type to send all N words?
Input
Input begins with an integer T, the number of test cases. For each test case, there is first a line containing the integer N. Then,N lines follow, each containing
a word to send in the order you wish to send them.
Output
For the ith test case, print a line containing "Case #i: " followed by the minimum number of characters you need to type in your text message.
Constraints
1 ≤ T ≤ 100
1 ≤ N ≤ 100,000
The N words will have a total length of no more than 1,000,000 characters.
The words are made up of only lower-case alphabetic characters.
The words are pairwise distinct.
NOTE: The input file is about 10⑵0MB.
Explanation of Sample
In the first test case, you will write "h", "he", "l", "hil", "hill", for a total of 1 + 2 + 1 + 3 + 4 = 11 characters.
Example input ・
Example output ・
Case #1: 11
Case #2: 15
Case #3: 11
Case #4: 9
Case #5: 3
解题思路：
使用字典树可以求出该字符串是不是和其他字符串有公共前缀。具体看代码
题目代码：
#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <cctype>
#include <algorithm>
#define eps 1e⑴0
#define pi acos(⑴.0)
#define inf 107374182
#define inf64 1152921504606846976
#define lc l,m,tr<<1
#define rc m + 1,r,tr<<1|1
#define zero(a) fabs(a)<eps
#define iabs(x) ((x) > 0 ? (x) : -(x))
#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))
#define clearall(A, X) memset(A, X, sizeof(A))
#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))
#define memcopyall(A, X) memcpy(A , X ,sizeof(X))
#define max( x, y ) ( ((x) > (y)) ? (x) : (y) )
#define min( x, y ) ( ((x) < (y)) ? (x) : (y) )
using namespace std;
struct node
{
int p[27];
int cnt;
bool alive;
} treenode[1000005];
int ans,nodecnt;
char s[1000005];
void insertto()
{
int len=strlen(s),p=0;
ans++;
if(treenode[p].p[s[0]-'a']==0)
{
clearall(treenode[nodecnt].p,0);
treenode[nodecnt].cnt=0;
treenode[nodecnt].alive=false;
treenode[p].p[s[0]-'a']=nodecnt++;
}
p=treenode[p].p[s[0]-'a'];
treenode[p].cnt++;
for(int i=1; i<len; i++)
{
if(treenode[p].p[s[i]-'a']==0)
{
clearall(treenode[nodecnt].p,0);
treenode[nodecnt].cnt=0;
treenode[nodecnt].alive=false;
treenode[p].p[s[i]-'a']=nodecnt++;
}
if(treenode[p].cnt==1&&treenode[treenode[p].p[s[i]-'a']].cnt==0)
{
}
else ans++;
treenode[treenode[p].p[s[i]-'a']].cnt++;
p= treenode[p].p[s[i]-'a'];
}
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.txt","w",stdout);
int t,case1=1;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
clearall(treenode[0].p,0);
treenode[0].cnt=0;
treenode[0].alive=false;
nodecnt=1;
ans=0;
int n;
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%s",s);
insertto();
//printf("%d
",ans);
}
printf("Case #%d: %d
",case1++,ans);
}
}
return 0;
}
题目终究测试数据：
链接: http://pan.baidu.com/s/1o6oiwY2
密码: 1dgp