badtraveller

算法题，逆出来大概是这样：

#include <cstdlib>
#include <iostream>
#include <queue>
#include <sstream>
#include <vector>
 
using namespace std;
 
int key1[40] = {3, 4, 1, 1, 8, 7, 4, 5, 2,  0, 1, 1, 2,  9, 8, 1, 2, 5, 2, 7,
                3, 1, 9, 6, 3, 8, 2, 5, 10, 2, 3, 7, 10, 6, 9, 8, 7, 8, 1, 0};
 
vector<string> tab1 = {"cnss{", "Never", "gonna", /* ... */};
 
struct Edge {
    int to;
    int weight;
};
 
struct Element { // maybe 0x40
    int index;
    vector<Edge> edges;
    std::string str;
    // other fields
};
Element arr[11];
 
void init() {
 
    for (int i = 0; i <= 10; ++i) {
        arr[i].index = i;
        arr[i].str = tab1[i];
    }
 
    for (int j = 0; j <= 38; j += 3) {
        int u = key1[j];
        int v = key1[j + 1];
        int w = key1[j + 2];
 
        addEdge(arr[u], v, w);
        addEdge(arr[v], u, w);
    }
}
 
int main() {
    string input;
    cin >> input;
    stringstream ss(input);
    int nowEdge = 0, sum = 0;
    string i;
    bool flag = false;
    while (getline(ss, i, '_')) {
 
        // 查找i在tab中的位置
        int index = -1;
        for (auto j : tab1) {
            index++;
            if (j == i) {
                break;
            }
        }
        if (index == -1) {
            cout << "Wrong!" << endl;
            return 0;
        }
        if (index) {
            for (auto j : arr[nowEdge].edges) {
 
                if (j.to == index) {
                    nowEdge = j.to;
                    sum += j.weight;
                    flag = true;
                    break;
                }
            }
            if (!flag) {
                cout << "Wrong!" << endl;
                return 0;
            }
        }
    }
}

string 24byte+

这里用的是临接表求最短路（唯一）