题目

What does our future hold? How will the universe meet its end? Over the whole human history, controversy on whether the universe ends in fire or ice has never come to an end. But now, with the second law of thermodynamics, maybe we can conclude that the universe ends in not a giant explosion, but a whimper.

After 10^377 years, this universe is approaching its end. Everything goes together, becoming a black hole, and nothing is left except some dying black dwarf stars which are the only places for possibly remaining civilizations to feed on. Now you are the God and only Sylvie is lingering on with her last breath of life in this universe created by you. With mercy, you are affected by her strong will to survive, so you decide, if she solves the problem you present her, you will ship her to the next universe you create where there are still colours.

So you whispered by her ear: ‘Every black dwarf star has energy, and a string of letters with it. If you live long enough to collect the largest amount of energy from those black dwarf stars, keeping on collecting strings of letters of each star you pass, combining them in turn along the journey, you will find the encoded answer therein. Decode it by finding the sequence of letters with the most front alphabetic order among permutations of the original without changing circular order of its letters. For example, if you pass three stars with string “b”, “cd” and “a” respectively, and combines them to get “bcda”, you can then decode it and get the answer “abcd”.’

Sylvie dies as soon as her energy runs out, and when arriving at a star, the road carrying her there takes certain amount of her energy and disappears, but she collects all the energy of that star which is, sadly, strictly nonrenewable.

There are* N* black dwarf stars and M directed roads between them in the universe and Sylvie is now on star S1 with roads leading to other stars and the energy costs along these roads are given. Every time Sylvie arrives at a planet, the string of the planet is collected again.

Because of the incredible deficiency of energy in the dying universe, even one spark of thought will take billions of years within which Sylvie will die for sure together with the universe. So Sylvie has to get the right answer quickly.

Since you are curious and almighty, you think of the answer immediately after having invented the problem. And in the meanwhile, the idea that putting your wisdom down in form of an ancient language invented by human beings for their computers has come to your mind for fun. Try doing it now!

Input Specification:

The first line are two integers N and M (N>0,M>0,N<=20 M<=300) and the second line, N integers, of which the ith integer describes the energy of the ith black dwarf star. M lines with three integers x,y,e are following, meaning star Sx has one directed road leading to star Sy costing e energy on the road. Finally there are N lines, of which each is a string of letters – the string can be found on the ith star (at the beginning, Sylvie has the energy of the first black dwarf star).

Output Specification:

Altogether two lines, of which the first line is the string of letters you have decoded, while the second, the amount of energy Sylvie collected. There is only one path to collect the largest amount of energy.

Sample Input:

4 5
10 10 10 20
1 2 3
1 3 5
2 4 10
3 4 5
1 4 1
b
f
cd
a

Sample Output:

abcd
30

基本思路

这题可以分为两个子问题：
第一个问题用求一条能获得最大能量的最优路径，跟普通的图的深搜问题不大一样，这里的顶点只要能量足够可以重复去访问，用回溯法解决。
第二个问题是循环字符串的最小表示法（不知道该算法的可以先了解一下），可以套用固定模板。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
//一个黑矮星对应一个结点
struct Star {int power;string letters;bool accessed;
};
Star stars[21];
int cost[21][21];
int n, m;//结点数，边数
int maxEnergyGot;//能获得的最大能量
vector<int> maxSeq;//能获得最大能量的路径void dfs(int start, int energyGot, vector<int> v) {for (int i = 1; i <= n; i++) {//如果顶点start到顶点i有路径 并且 携带的能量足够走这条路路径if (cost[start][i] && energyGot >= cost[start][i]) {//bool changed = false;//判断顶点i是不是在这个dfs中被访问的if (!stars[i].accessed) {//如果顶点i没有被访问过changed = true;stars[i].accessed = true;energyGot += stars[i].power - cost[start][i];} else{//如果顶点i被访问过energyGot += -cost[start][i];}v.push_back(i);//如果临时能量大于最大能量，更新最大能量和最优路径if (energyGot > maxEnergyGot) {maxEnergyGot = energyGot;maxSeq.assign(v.begin(), v.end());//数组拷贝}//对顶点i、临时能量energyGot、临时路径v进行dfsdfs(i, energyGot, v);//v.pop_back();if (changed) {stars[i].accessed = false;energyGot -= stars[i].power - cost[start][i];}else{energyGot -= -cost[start][i];}}}
}int solve(string s){int i=0,j=1;string t=s+s;while (i<s.length() && j<s.length()){int k=0;while(k<s.length() &&t[i+k]==t[j+k]) k++;if(k==s.length()) break;if(t[i+k]>t[j+k]) i=i+k+1;else j=j+k+1;if(i==j) i++;}return  min(i,j);
}int main() {//读入cin >> n >> m;for (int i = 1; i <= n; i++)cin >> stars[i].power;for (int i = 1; i <= m; i++) {int a, b, e;cin >> a >> b >> e;cost[a][b] = e;}for (int i = 1; i <= n; i++)cin >> stars[i].letters;//把起点的能量赋值给最大能量，把起点放入临时路径v，并把临时路径v拷贝给最优路径maxSeqmaxEnergyGot = stars[1].power;vector<int> v;//临时路径vv.push_back(1);maxSeq.assign(v.begin(), v.end());//从起点、起点的能量、临时路径v开始dfsdfs(1, stars[1].power, v);//取出最优路径上每个顶点所携带的字符串letters，把他们拼接成字符串res//最小字典序的循环同构串的求解string res;for(auto i:maxSeq){res+=stars[i].letters;}int pos= solve(res);//输出for(int i=pos;i<res.length();i++)cout<<res[i];for(int i=0;i<pos;i++)cout<<res[i];cout<<endl<<maxEnergyGot<<endl;return 0;
}

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