uva 10602 - Editor Nottoobad（贪心）

Problem C

EDITOR NOTTOOBAD

 

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	this this
Say “delete the last symbol”	0	this thi
Press "n"	1	this thin
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 winner J, 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 editor Nottoobad. Next N lines contain the words in that minimum order. If there are several solutions, you can output one of them.

 

 

 

 

 

Sample Input	Sample Output
3 3 this thin thing 4 popcorn apple apricote plum 2 hello hello	6 this thin thing 21 popcorn plum apricote apple 5 hello hello

 

Problem source: Russian National Team Olympiad 2000

Problem author: Andrew Stankevich

Problem translation: Dmytro Chernysh

Problem solution: Andrew Stankevich, Dmytro Chernysh

 开始想先预处理出任意两个字符串间的最短编辑距离，然后求从起点经过所有顶点的最短距离。转化到这个问题后，不知道要怎么做了。想贪心的每次找离当前点距离最近的点作为下一个点，但发现是错的。通过举了一个反例后，突然想到，其实公共前缀长才是节省步骤的关键。每次尽量把最长的公共前缀保留下来，利用这一点正确的贪心算法就出来了。

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#include<stack>
#include<iostream>
#include<queue>
using namespace std;
const int maxn = 100 + 5;
typedef pair<int, string> P;

char s[maxn][maxn];
int common[maxn][maxn];
int vis[maxn];
vector<int> ans;

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int n;
        scanf("%d", &n);
        for(int i = 0;i < n;i++)
            scanf("%s", s[i]);
        for(int i = 0;i < n;i++){
            for(int j = i+1;j < n;j++){
                int len = min(strlen(s[i]), strlen(s[j]));
                int cnt = 0;
                for(int k = 0;k < len;k++){
                    if(s[i][k]==s[j][k]){
                        cnt++;
                    }
                    else
                        break;
                }
                common[i][j] = common[j][i] = cnt;
            }
        }
        int step = strlen(s[0]);
        int now = 0;
        memset(vis, 0, sizeof vis);
        ans.clear();
        ans.push_back(0);
        for(int k = 0;k < n-1;k++){
            int Max = -1, id;
            for(int i = 1;i < n;i++){
                if(!vis[i] && common[now][i] > Max){
                    Max = common[now][i];
                    id = i;
                }
            }
            vis[id] = 1;
            step += strlen(s[id])-Max;
            ans.push_back(id);
            now = id;
        }
        printf("%d\n", step);
        for(int i = 0;i < n;i++){
            printf("%s\n", s[ans[i]]);
        }
    }
    return 0;
}