[F] Teacher's Problem(处理大数时，优化很重要)

1、http://acm.nbut.edu.cn/Problem/view.xhtml?id=1546

2、题目大意：

有n个字符串，每个字符串都有一个价值，如果第i个字符串是第j个字符串的前缀，那么第i个字符串就是第j个字符串的父亲，这样就会组成一棵树，子节点的价值都更新到父节点的价值上，然后按顺序输出即可

3、题目：

[F] Teacher's Problem

时间限制: 2000 ms　内存限制: 262144 K       

问题描述

One day my teacher gave me a problem to solve. I was given N unique names andeach name has a value. Name is a string consisting of lowercase Latin letters withlength less than 10. Value is a positive integer less than 100. Initially these namesare sorted in lexicographic order. Now if the j-th name is the prefix of the i-th nameand has the max length, j is thefather of i. Then numbers 1 to N form a tree. Fromleafto root, each son's value should be added to father's value. After I added thesevalues, my teacher ask me to show allthese values.

Can you help me with the problem?

输入

First line there is an integer T (T < 20) following T test cases.
For each test case, first line is an integer N (0 < N <= 10^5), second line is N names, third line is N values.
Names and values are separate by a space. There is no space at the end of line.
Names are all unique and are given in Lexicographic order.           

输出

For each test case, please output one line with N integers. The i-th integer is the i-th value after added.
Please do not output a space at the end of line.           

样例输入

2
3
a ab ac
5 10 15
5
a ab abb ac b
5 10 100 15 6

样例输出

30 10 15
130 110 100 15 6

提示

Huge input, scanf is recommended.

来源

zzxchavo @HEU 

4、Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 100005
char s[N][12];
int w[N];
int check(int a,int b)
{
    int len=strlen(s[a]);
    for(int i=0;i<len;i++)
    {
        if(s[a][i]!=s[b][i])
        return 0;
    }
    return 1;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s[i]);
        }
        for(int i=1;i<=n;i++)
        scanf("%d",&w[i]);
        for(int i=1;i<n;i++)
        {
            int sum=0;
            sum+=w[i];
            for(int j=i+1;j<=n;j++)
            {
                if(check(i,j))
                sum+=w[j];
                else
                break;
            }
            printf("%d ",sum);
        }
        printf("%d\n",w[n]);
    }
    return 0;
}