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

Project managers, such as the UNIX utility make,are used to maintain large software projects made up frommany components.Users write a project file specifying which components(called tasks) dependon others and the project manager canautomatically update the components in the correct order.

Write a program that reads a project fileand outputs the order in which the tasks should be performed.

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.

For simplicity we represent each task by an integer numberfrom 1,2,...,N (where N is the total number of tasks).The first line of input specifies the number N of tasksand the number M of rules,such that N ≤ 100, M ≤ 100.

The rest of the input consists of M rules, one in each line,specifying dependencies using the following syntax:

T₀      k      T₁     T₂      ...     T_k

This rule means that task number T₀ depends onk tasks T₁, T₂, ..., T_k(we say that task T₀ is the targetand T₁,...,T_k are dependents).

Note that tasks numbers are separated by single spacesand that rules end with a newline.Rules can appear in any order, but each task can appear as targetonly once.

Your program can assume that there are no circular dependencies in therules, i.e. no task depends directly or indirectly on itself.

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 should be a single linewith the permutation of the tasks 1 ... N to be performed,ordered by dependencies (i.e. no task should appear before othersthat it depends on).

To avoid ambiguity in the output, tasks that do not depend on eachother should be ordered by their number (lower numbers first).

Sample Input 

15 43 2 1 52 2 5 34 1 35 1 1

Sample Output 

1 5 3 2 4

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題目要求一個字典順序的拓樸排序，將 queue 改成 priority_queue 即可。

#include <stdio.h>
#include <queue>
#include <functional>
#include <vector>
using namespace std;

int main() {
    priority_queue<int, vector<int>, greater<int> > pQ;
    vector<int> g[105];
    int t, M, N, indeg[105];
    int i, j, x, y, z;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &N, &M);
        for(i = 0; i <= N; i++)
            g[i].clear(), indeg[i] = 0;
        while(M--) {
            scanf("%d %d", &x, &z);
            while(z--) {
                scanf("%d", &y);
                g[y].push_back(x);
                indeg[x]++;
            }
        }
        for(i = 1; i <= N; i++)
            if(indeg[i] == 0)
                pQ.push(i);
        int tn, flag = 0;
        while(!pQ.empty()) {
            if(flag)    putchar(' ');
            flag = 1;
            tn = pQ.top(), pQ.pop();
            printf("%d", tn);
            for(vector<int>::iterator it = g[tn].begin();
                it != g[tn].end(); it++)
                if(--indeg[*it] == 0)
                    pQ.push(*it);
        }
        puts("");
        if(t)
            puts("");
    }
    return 0;
}