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

"Gossipy gossiper, what did you get of so much gossiping?
You set your soul into hell. Behold your great achievement!"
M.A. Hevia. Mermuradora.

The Problem

Gossiping mechanism is quite simple, but effective. Everything starts with a master gossiper (most probably a she), who hears, witnesses, or makes up some extraordinary news. Whenever she meets another person, she tells him the news. As soon as these people meet others, they tell them the news. Soon, everybody is informed of the news, and the process finishes.

Gossipers meet each other daily at the same hours. We have information on when every pair of gossipers meet. For simplicity, days are divided into 100 instants. The first day goes from 0 to 99, the second from 100 to 199, and so on.

Suppose the gossiping process starts at time 0. When will it finish? That is, when will all the people be informed of the news?

The Input

The first line of the input contains an integer N, indicating the number of test cases.

For each test case, the first line contains two integers M and K. M indicates the number of people in this case (numbered from 1 to M). The master gossiper is always number 1. K indicates the number of pairs of people who meet.

Next, we have 2K lines (two lines for each pair of people who meet). The first line of each pair contains three integers, G1, G2 and V. This line means that gossipers G1 and G2 meet V times a day. The second line contains V numbers from 1 to 99, that indicate the instants when G1 and G2 meet daily.

For example, the pair:

3 7 2
28 88

means: person 3 and person 7 meet twice a day, at instants 28 and 88 everyday (that is, at 28, 88, 128, 188, 228, 288, ...).

You can assume that M ≤ 20, K ≤ M*M, and V ≤ 12 .

The Output

For each test case, the output should consist of a single integer F in one line, indicating the instant when the process finishes. If the process does not finish (for example, there is some isolated person who will never be informed), the result should be -1.

Sample Input

4
4 3
1 2 1
60
2 3 1
40
4 3 1
30
3 2
1 2 1
42
2 3 1
42
5 5
1 3 2
60 70
1 4 3
20 22 24
4 5 1
10
3 5 2
12 80
3 2 2
55 78
8 0

Sample Output

230
42
80
-1

*Warning: "gossips" have nothing to do with "ghost ships".

---

題目描述：

八卦傳遞，每一天可以劃分成 100 個時槽 [0, 99]，
問從編號 1 (從時刻 0 開始)將八卦散布於全部人都知道的最少時間為何。

給予 x y 之間會在哪些時刻交換訊息，逼不得已的情況下必須到下一天進行。

題目解法：

SSSP，單源最短路徑。

#include <stdio.h>
#include <vector>
#include <queue>
using namespace std;
struct edge {
    int to, time;
    edge(int a, int b):
        to(a), time(b) {}
};
vector<edge> g[25];
void spfa(int n) {
    int dp[25][105], inq[25][105];
    int i, j, k;
    int x, t, y, w;
    for(i = 1; i <= n; i++)
        for(j = 0; j < 100; j++)
            dp[i][j] = 0xfffffff, inq[i][j] = 0;
    dp[1][0] = 0;
    queue<int> X, TIME;
    X.push(1), TIME.push(0);
    while(!X.empty()) {
        x = X.front(), X.pop();
        t = TIME.front(), TIME.pop();
        inq[x][t] = 0;
        for(i = 0; i < g[x].size(); i++) {
            y = g[x][i].to, w = g[x][i].time;
            int gap;
            if(w < t)    gap = 100 - t + w;
            else        gap = w - t;
            if(dp[y][w] > dp[x][t] + gap) {
                dp[y][w] = dp[x][t] + gap;
                if(inq[y][w] == 0) {
                    inq[y][w] = 1;
                    X.push(y), TIME.push(w);
                }
            }
        }
    }
    int ret = 0;
    for(i = 1; i <= n; i++) {
        int temp = 0xfffffff;
        for(j = 0; j < 100; j++)
            temp = min(temp, dp[i][j]);
        ret = max(ret, temp);
        if(temp == 0xfffffff) {
            ret = -1;
            break;
        }
    }
    printf("%d\n", ret);
}
int main() {
    int testcase;
    int n, m, x, y, t, time;
    int i, j, k;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &n, &m);
        for(i = 1; i <= n; i++)
            g[i].clear();
        for(i = 0; i < m; i++) {
            scanf("%d %d %d", &x, &y, &t);
            while(t--) {
                scanf("%d", &time);
                g[x].push_back(edge(y, time));
                g[y].push_back(edge(x, time));
            }
        }
        spfa(n);
    }
    return 0;
}