Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

By 2100, space travel will be a reality for the Milky Way (Solar Galaxy) residents. The Interstar Transport Travel agency operates scheduled direct space transports (flights) between some of the most popular planet destinations in the Milky Way. The cost of these scheduled direct transports are predetermined in Galaro (Galaxy Currency unit) and are published in many different languages. For travel to planets that is not on the schedule, there are slower, yet free, space flights from the closest planet that is on the direct transport list. To help travelers plan their itinerary, the Interstar Transport wants to offer a mobile application that can find the best traveling option, based on the total cost of the direct transports on the itinerary. Given the starting and destination planets on the itinerary, find the sequence of direct transports that has the lowest total traveling cost. Output all the planets in sequence that one must pass through on this best route. If two or more routes exist with the same cost, then the route that goes through the least number of intermediate planets is considered a better route. There will always exist a unique best route for any of the given test cases.

Technical Specification

1. The number of planets on the direct transport list is at most s, 1s26. The planets are labeled using capital letters of the English alphabets, i.e., ``A", ``B", ``C", ..., ``Z", in no particular order.
2. The Interstar Transport operates at most p, 1p200, direct transports between planets. There is at most one (could be none) direct transport between any two distinct planets.
3. The cost of any transport is given as a natural number less than or equal to 100 Galaros.

Input 

The first line of the input file contains two integers, s and p, separated by a space. The next p lines each contains two letters, e_i and e_j, followed by a natural number, d_ij, indicating that there exists a direct transport between planets e_i and e_j with a cost of d_ij. The next line contains an integer n20, indicating the number of queries to follow. For each of the next n lines, each line contains two letters e_k and e_m, indicating a user is looking for a best (lowest cost) way to get from planet e_k to planet e_m.

Output 

For each of the n queries in the input, output on one line the best route to take, in the sequence of starting planet, the intermediate planets in sequence along the route and the destination planet; all separated by one blank space.

Sample Input 

5 7 
A B 1
B C 2
C D 3
D E 2
E A 1
A D 3
A C 4
3 
B D 
A D 
E C

Sample Output 

B A D 
A D 
E A B C

---

單純地找最短路徑後打印出結果。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <queue>
using namespace std;
int g[30][30] = {};
void spfa(int st, int ed) {
    int dist[30][30], pre[30][30];
    int inq[30][30] = {};
    int i, j;
    for(i = 0; i < 26; i++)
        for(j = 0; j < 26; j++)
            dist[i][j] = 0xfffffff;
    dist[st][0] = 0;
    queue<int> X, P;
    X.push(st), P.push(0);
    while(!X.empty()) {
        int x = X.front();
        int p = P.front();
        X.pop(), P.pop();
        inq[x][p] = 0;
        for(i = 0; i < 26; i++) {
            if(dist[i][p+1] > dist[x][p] + g[x][i]) {
                dist[i][p+1] = dist[x][p] + g[x][i];
                pre[i][p+1] = x;
                if(inq[i][p+1] == 0) {
                    inq[i][p+1] = 1;
                    X.push(i), P.push(p+1);
                }
            }
        }
    }
    int mn = dist[ed][0], x = ed, p = 0;
    for(i = 0; i < 26; i++)
        if(dist[ed][i] < mn)
            mn = dist[ed][i], p = i;
    printf("%c", x+'A');
    while(x != st) {
        x = pre[x][p];
        p--;
       // printf("%d\n", p);
        printf(" %c", x+'A');
    }
    puts("");
}
int main() {
    int n, m, q;
    int i, j, k;
    while(scanf("%d %d", &n, &m) == 2) {
        for(i = 0; i < 26; i++) {
            for(j = 0; j < 26; j++)
                g[i][j] = 0xfffffff;
            g[i][i] = 0;
        }
        char s1[5], s2[5];
        int v;
        while(m--) {
            scanf("%s %s %d", s1, s2, &v);
            g[s1[0]-'A'][s2[0]-'A'] = v;
            g[s2[0]-'A'][s1[0]-'A'] = v;
        }
        scanf("%d", &q);
        while(q--) {
            scanf("%s %s", s1, s2);
            spfa(s2[0]-'A', s1[0]-'A');
        }
    }
    return 0;
}