圖論：拓撲排序
來源：程序員人生 發布時間：2015-01-17 09:49:28 閱讀次數：4563次
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，輸出順序任意，所以DFS做法可以求得。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
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=cnt⑴;i>=0;i--)
printf(i==0?"%d
":"%d ",toposort[i]);
}
return 0;
}
HDU 1285
肯定比賽名次
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個比賽隊（1<=N<=500），編號順次為1，2，3，。。。。，N進行比賽，比賽結束后，裁判委員會要將所有參賽隊伍從前往后順次排名，但現在裁判委員會不能直接取得每一個隊的比賽成績，只知道每場比賽的結果，即P1贏P2，用P1，P2表示，排名時P1在P2之前?，F在請你編程序肯定排名。
Input
輸入有若干組，每組中的第1行動2個數N（1<=N<=500），M；其中N表示隊伍的個數，M表示接著有M行的輸入數據。接下來的M行數據中，每行也有兩個整數P1，P2表示即P1隊贏了P2隊。
Output
給出1個符合要求的排名。輸出時隊伍號之間有空格，最后1名后面沒有空格。
其他說明：符合條件的排名可能不是唯1的，此時要求輸出時編號小的隊伍在前；輸入數據保證是正確的，即輸入數據確保1定能有1個符合要求的排名。
Sample Input
Sample Output
字典序最小的。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
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<n)
{
for(i=1;i<=n;i++)
{
if(!vis[i]&&!in[i])
{
que[++k]=i;
vis[i]=1;
break;
}
}
for(int j=1;j<=n;j++)
{
if(mp[i][j]) --in[j];
}
}
}
void output()
{
for(int i=1;i<=k;i++)
printf(i==k?"%d
":"%d ",que[i]);
}
int main()
{
int x,y;
std::ios::sync_with_stdio(false);
while(~scanf("%d%d",&n,&m))
{
CLEAR(in,0);
CLEAR(mp,0);
CLEAR(vis,0);
while(m--)
{
scanf("%d%d",&x,&y);
if(!mp[x][y])
{
mp[x][y]=1;
in[y]++;
}
}
toposort();
output();
}
return 0;
}
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,..., N⑴). 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
判斷能不能拓撲排序。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
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;t++)
{
for(i=0;i<n;i++)
{
if(!vis[i]&&!in[i])
{
vis[i]=1;
in[i]--;
k++;
for(int j=0;j<n;j++)
if(mp[i][j]) in[j]--;
break;
}
}
}
// cout<<"2333 "<<k<<endl;
return k==n?1:0;
}
int main()
{
int x,y;
std::ios::sync_with_stdio(false);
while(cin>>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"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
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