HDU 5154 Harry and Magical Computer（拓扑排序模板）

Problem Description：

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

 

Input：

There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n

 

Output：

Output one line for each test case. 
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).

 

Sample Input：

3 2

3 1

2 1

3 3

3 2

2 1

1 3

 

Sample Output：

YES

NO

题意：有n个任务需要全部完成，但是现在有m个条件，必须在满足这m个条件的同时完成n个任务，这m个条件中每组的两个任务（a，b），任务b必须在任务a完成之前完成，问能否完成所有的任务。

 

拓扑排序（并输出该序列）：1.从有向图中找一个没有前趋的结点v（入度为0），若v不存在，则表明不可进行拓扑排序（有环路），结束；

2.输出v；

3.将v从图中删除，同时删除与v有关的所有边；

4.若图中所有点均已输出，则结束（成功排序），否则转1继续进行。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

typedef long long LL;

struct node
{
    int v, next;
}no[N];
int Head[N], ru[N], vis[N], k;

void Init()
{
    memset(Head, -1, sizeof(Head));
    memset(ru, 0, sizeof(ru)); ///记录每个结点的入度个数
    memset(vis, 0, sizeof(vis)); ///标记该点是否被查找

    k = 0;
}

void Add(int a, int b)
{
    no[k].v = b;
    no[k].next = Head[a];

    Head[a] = k++;
}

int main ()
{
    int n, m, flag, i, num, Min, v, a, b;

    while (scanf("%d%d", &n, &m) != EOF)
    {
        Init();
        num = 0;
        flag = 1;

        while (m--)
        {
            scanf("%d%d", &a, &b);

            ru[a]++;
            Add(b, a);
        }

        while (num < n)
        {
            Min = INF;

            for (i = 1; i <= n; i++)
            {
                if (!ru[i] && i < Min && !vis[i]) ///查找入度为0的最小结点
                    Min = i;
            }

            if (Min == INF) ///没有查找到入度为0的结点，排序失败，结束
            {
                flag = 0;
                break;
            }

            vis[Min] = 1; ///将查找过的入度为0的点标记（即在图中删除该点）

            for (i = Head[Min]; i != -1; i = no[i].next)
            {
                v = no[i].v;
                ru[v]--; ///与之相连的边也需要删除
            }

            num++; ///num记录成功结点的个数
        }

        if (flag == 1) printf("YES
");
        else printf("NO
");
    }

    return 0;
}