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

A set of laboratory mice is being trained to escape a maze. The maze is made up of cells, and each cell isconnected to some other cells. However, there are obstacles in the passage betweencells and therefore there is a time penalty to overcome the passageAlso, some passages allow mice to go one-way,but not the other way round.

Suppose that all mice are now trained and, when placed in an arbitrarycell in the maze, take a path that leads them to the exit cell in minimumtime.

We are going to conduct the following experiment: a mouse is placed in each cell ofthe maze and a count-down timer is started.When the timer stops we count the number of mice out of the maze.

Write a program that, given a description of the maze and the time limit,predicts the number of mice that will exit the maze. Assumethat there are no bottlenecks is the maze, i.e. that all cells haveroom for an arbitrary number of mice.

Input 

The input begins with a single positive integer on a line by itself indicatingthe number of the cases following, each of them as described below.This line is followed by a blank line, and there is also a blank line betweentwo consecutive inputs.

The maze cells are numbered 1,2,...,N, where N is the total numberof cells. You can assume that N ≤ 100.

The first three input lines contain N , the number of cells in themaze, E, the number of the exit cell,and the starting value T for the count-down timer (in some arbitrary time unit).

The fourth line contains the number M of connections in the maze, and is followed by M lines,each specifying a connectionwith three integer numbers: two cell numbers a and b(in the range 1,..,N) and the number of time unitsit takes to travel from a to b.

Notice that each connection is one-way, i.e., the mice can't travel from b to aunless there is another line specifying that passage.Notice also that the time requiredto travel in each direction might be different.

Output 

For each test case, the output must follow the description below.The outputs of two consecutive cases will be separated by a blank line.

The output consists of a single line with the number of mice that reached the exit cell E in atmost T time units.

Sample Input 

14 2 181 2 11 3 12 1 12 4 13 1 13 4 14 2 14 3 1

Sample Output 

3

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給迷宮兩點之間會有障礙物要清除(雙向)，接著在限定時間內到達終點的老鼠有幾隻？
假設每個點都會有一隻老鼠。

可以選擇使用 dijkstra 或者 spfa 都可以，對於終點反過來當最短路的起點即可。

#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
struct edge {
    int to, v;
    edge(int a, int b):
        to(a), v(b) {}
};
int main() {
    int testcase;
    int n, m, st, time;
    int i, j , k, x, y, v;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d %d", &n, &st, &time, &m);
        vector<edge> g[105];
        while(m--) {
            scanf("%d %d %d", &x, &y, &v);
            g[y].push_back(edge(x, v));
        }
        int dist[105], inq[105] = {};
        memset(dist, 63, sizeof(dist));
        dist[st] = 0;
        queue<int> Q;
        Q.push(st);
        while(!Q.empty()) {
            int tn = Q.front();
            Q.pop(), inq[tn] = 0;
            for(vector<edge>::iterator it = g[tn].begin();
                it != g[tn].end(); it++) {
                if(dist[it->to] > dist[tn] + it->v) {
                    dist[it->to] = dist[tn] + it->v;
                    if(inq[it->to] == 0) {
                        inq[it->to] = 1;
                        Q.push(it->to);
                    }
                }
            }
        }
        int ret = 0;
        for(i = 0; i <= n; i++ )
            if(dist[i] <= time)
                ret++;
        printf("%d\n", ret);
        if(testcase)
            puts("");
    }
    return 0;
}