UVa 10602 - Editor Nottoobad

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=1543

类型：贪心

原题：

Company Macrohard has released it’s new version of editor Nottoobad, which can understand a few voice commands. Unfortunately, there are only two voice commands that it can understand – “repeat the last word”, “delete the last symbol”. However, when one uses “repeat the last word” the editor inserts a blank that separates the words. But the company claims that it is possible to type much faster – simply by less number of presses. For example, such a phrase like “this thin thing” requires only 6 presses of the keyboard.

Action	Number of presses	Content of the document
Press "this"	4	This
Say “repeat the last word”	0	thisthis
Say “delete the last symbol”	0	this thi
Press "n"	1	thisthin
Say “repeat the last word”	0	this thin thin
Press "g"	1	this thin thing

In order to increase the popularity of it’s product the company decided to organize a contest where the winner will be a person who types a given number of words with minimum number of presses. Moreover, the first word must be typed first, and all the others can be typed in arbitrary order. So, if words “apple”, “plum” and “apricote” must be typed, the word “apple” must be typed first, and the words “plum” and “apricote” can be switched. And the most important for you – you are going to take part in the contest and you have a good friend in the company, who told you the word which will be used in the contest. You want be a winnerJ, so you have to write a program which finds the order of the words, where the number of presses will be minimum.

Input

The first line of the input contains the T (1≤T≤15) the number of test cases. Then T test cases follow. The first line of each test contains a number N (1≤N≤100) – the number of words that must be pressed. Next N lines contain words – sequences of small Latin letters, not longer than 100 symbols. Remember that the first word must be pressed first!

Output

The first line of the output contains number X - the minimum number of presses, which one has to do in order to type all the words using editorNottoobad. Next N lines contain the words in that minimum order. If there are several solutions, you can output one of them.

样例输入：

3

3

this

thin

thing

4

popcorn

apple

apricote

plum

2

hello

hello

样例输出;

6

this

thin

thing

21

popcorn

plum

apricote

apple

5

hello

hello

分析与总结：

贪心题，每次直接遍历选择一个可以减少输入次数最少的单词即可。

代码：

/*
 * UVa: 10602 - Editor Nottoobad
 * Result: Accept
 * Time: 0.020s
 * Author: D_Double
 */
#include<iostream>  
#include<cstdio>  
#include<string>  
#include<algorithm>  
#include<cstring>
#define MAXN 120  
using namespace std;  

struct Node{
    int no;
    char word[110];
    friend bool operator < (const Node &a, const Node &b){
        return a.no < b.no;
    }
}arr[MAXN];


int main(){
    int T, N;
    char str[110];
    scanf("%d", &T);
    while(T--){
        scanf("%d",&N);
        for(int i=0; i<N; ++i){
            scanf("%s",arr[i].word);
            arr[i].no = -1;
        }
        arr[0].no = 1;
        int ans=strlen(arr[0].word);
        int rank=1, last=0;

        while(rank < N){
            int t=-1, cnt=0;
            for(int i=0; i<N; ++i)if(arr[i].no==-1){
                if(t==-1){
                    t=i;
                    if(arr[i].word[0]==arr[last].word[0]){
                        for(int j=0; j<strlen(arr[i].word)&&j<strlen(arr[last].word); ++j){
                            if(arr[i].word[j]!=arr[last].word[j]) break;
                            else ++cnt;
                        }
                    }
                }
                else{            
                    int tmp=0;
                    for(int j=0; j<strlen(arr[i].word)&&j<strlen(arr[last].word); ++j){
                        if(arr[i].word[j]!=arr[last].word[j]) break;
                        else ++tmp;
                    }
                    if(tmp > cnt){
                        t=i; cnt=tmp;
                    }
                }
            }
            ans += (strlen(arr[t].word) - cnt);
            arr[t].no = rank++;
            last = t;
        } 
        sort(arr, arr+N);
        printf("%d\n", ans);
        for(int i=0; i<N; ++i)
            puts(arr[i].word);
    }
    return 0;
}

—— 生命的意义，在于赋予它意义。
    
  原创http://blog.csdn.net/shuangde800，By  D_Double (转载请标明)