hdu 1285 拓扑排序（邻接表+邻接矩阵）

/*

	题意：  给N个比赛结果，输出拓扑排名

			



	YY：拓扑排序，初搞，最近练邻接表，感觉比邻接矩阵快的多

		两个都写了一下，竟差两倍之多，邻接表神物啊!!!



*/



//邻接表  	Accepted	1285	15MS	248K	1030 B	C++	kfinder

#include <iostream>

using namespace std;

int ind[505];		// indegree入度个数

int adj[250010];    //adjacency list邻接表位置值

int adj_next[250010];//邻接表下一指针

int tail[505];		//邻接表尾



int main()

{

	int n,m,i,j,a,b;

	while(scanf("%d%d",&n,&m)!=EOF)

	{

		for(i = 0; i <= n; i++) { 

			tail[i] = -1;

			adj[i] = -1;

			adj_next[i] = -1;

			ind[i] = 0;

		}

		for(i = 0; i < m; ++i)

		{

			scanf("%d%d",&a,&b);

			int x = tail[a],flag = 0;

			while(x != -1)  //判断是否重边

			{

				if(adj[x] == b){

					flag = 1;

					break;

				}

				x = adj_next[x];

			}

			if(!flag)//关联a的邻接表

			{

				adj[i] = b;

				adj_next[i] = tail[a];

				tail[a] = i;

				ind[b] ++;

			}

		}

		for(i = 1;i <= n; i++)//找n次

		{

			for(j = 1;j <= n;j++)//遍历

			{

				if(ind[j] == 0){//当入度为0时，说明靠前

					ind[j] = -1;//在下次寻找入度为0时跳过

					//

					if(i == 1)	 printf("%d",j);

					else		 printf(" %d",j);

					//

					for(int k = tail[j]; k != -1; k = adj_next[k])//邻接位置入度减一

					{

						ind[adj[k]] --;

					}

					break;

				}

			}

		}

		printf("\n");

	}

	return 0;

}





//邻接矩阵  Accepted	1285	31MS	1232K	759 B	C++	kfinder

#include <iostream>

using namespace std;

int map[505][505];

int indegree[505];

int n,m;



int main()

{

    int i,a,b;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        memset(map,0,sizeof(map));

        memset(indegree,0,sizeof(indegree));

        for(i = 0;i < m; i++)

        {

            scanf("%d %d",&a,&b);

            if(!map[a][b])

            {

                map[a][b] = 1;

                indegree[b]++; 

            }

        }

        for(i = 1;i <= n; i++)

        {

            for(int j = 1;j <= n; j++)

            {

                if(indegree[j] == 0){

                    indegree[j] = -1;

                    //

                    if(i == 1) printf("%d",j);

                    else printf(" %d",j);

                    //

                    for(int k = 1;k <= n; k++)

                    {

                        if(map[j][k] == 1){

                            map[j][k] = 0;

                            indegree[k] --;

                        }

                    }

                    break;

                }

            }

        }

        printf("\n");

    }

    return 0;

}