频道栏目
首页 > 程序开发 > 软件开发 > Java > 正文
拓扑排序入门
2014-12-04 10:57:48           
收藏   我要投稿

10305 - Ordering Tasks

John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.

Input

The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 andm. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0will finish the input.

Output

For each instance, print a line with n integers representing the tasks in a possible order of execution.

Sample Input

5 4
1 2
2 3
1 3
1 5
0 0

Sample Output

1 4 2 5 3
裸的,输出顺序任意,所以DFS做法可以求得。

#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=110;
int mp[maxn][maxn];
int toposort[maxn];
int vis[maxn];
int n,m,cnt;
void dfs(int x)
{
    vis[x]=1;
    for(int i=1;i<=n;i++)
        if(!vis[i]&&mp[x][i]&&i!=x)   dfs(i);
    toposort[cnt++]=x;
}
void solve()
{
    CLEAR(vis,0);
    for(int i=1;i<=n;i++)
       if(!vis[i])  dfs(i);
}
int main()
{
     int x,y;
     std::ios::sync_with_stdio(false);
     while(cin>>n>>m&&(n+m))
     {
         CLEAR(mp,0);
         while(m--)
         {
             cin>>x>>y;
             mp[x][y]=1;
         }
         cnt=0;
         solve();
         for(int i=cnt-1;i>=0;i--)
            printf(i==0?"%d\n":"%d ",toposort[i]);
     }
     return 0;
}
HDU 1285

确定比赛名次

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13115 Accepted Submission(s): 5264


Problem Description 有N个比赛队(1<=N<=500),编号依次为1,2,3,。。。。,N进行比赛,比赛结束后,裁判委员会要将所有参赛队伍从前往后依次排名,但现在裁判委员会不能直接获得每个队的比赛成绩,只知道每场比赛的结果,即P1赢P2,用P1,P2表示,排名时P1在P2之前。现在请你编程序确定排名。

Input 输入有若干组,每组中的第一行为二个数N(1<=N<=500),M;其中N表示队伍的个数,M表示接着有M行的输入数据。接下来的M行数据中,每行也有两个整数P1,P2表示即P1队赢了P2队。

Output 给出一个符合要求的排名。输出时队伍号之间有空格,最后一名后面没有空格。

其他说明:符合条件的排名可能不是唯一的,此时要求输出时编号小的队伍在前;输入数据保证是正确的,即输入数据确保一定能有一个符合要求的排名。

Sample Input
4 3
1 2
2 3
4 3

Sample Output
1 2 4 3

字典序最小的。。

#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=550;
int mp[maxn][maxn],vis[maxn];
int in[maxn],que[maxn];
int n,m,k;
void toposort()
{
    k=0;int i;
    while(k
HDU 3342

Legal or Not

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4619 Accepted Submission(s): 2106


Problem Description ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless,some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.
Input The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
Output For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".
Sample Input
3 2
0 1
1 2
2 2
0 1
1 0
0 0

Sample Output
YES
NO

判断能不能拓扑排序。

#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=110;
int mp[maxn][maxn],vis[maxn];
int in[maxn];
int n,m,k;
int toposort()
{
    k=0;int i;
    for(int t=0;t>n>>m&&(n+m))
     {
         CLEAR(mp,0);
         CLEAR(vis,0);
         CLEAR(in,0);
         while(m--)
         {
             cin>>x>>y;
             if(!mp[x][y])
             {
                 mp[x][y]=1;
                 in[y]++;
             }
         }
         if(toposort())   cout<<"YES"<

点击复制链接 与好友分享!回本站首页
相关TAG标签 拓扑
上一篇:设计模式的七大原则
下一篇:java-接口
相关文章
图文推荐
文章
推荐
点击排行

关于我们 | 联系我们 | 广告服务 | 投资合作 | 版权申明 | 在线帮助 | 网站地图 | 作品发布 | Vip技术培训 | 举报中心

版权所有: 红黑联盟--致力于做实用的IT技术学习网站