DZY Loves Topological Sorting (BC #35 hdu 5195 topsort+优先队列)

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 264 Accepted Submission(s): 63



Problem Description

A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge (u→v) from
vertex u to
vertex v,u comes
before v in
the ordering.

Now, DZY has a directed acyclic graph(DAG). You should find the lexicographically largest topological ordering after erasing at most k edges
from the graph.



Input

The input consists several test cases. (TestCase≤5)

The first line, three integers n,m,k(1≤n,m≤105,0≤k≤m).

Each of the next m lines
has two integers: u,v(u≠v,1≤u,v≤n),
representing a direct edge(u→v).



Output

For each test case, output the lexicographically largest topological ordering.



Sample Input

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

Sample Output

5 3 1 2 4
1 3 2

HintCase 1.
Erase the edge (2->3),(4->5).
And the lexicographically largest topological ordering is (5,3,1,2,4).

Source

BestCoder Round #35



Recommend

hujie | We have carefully selected several similar problems for you: 5197 5196 5193 5192 5189

题意：n个点m条有向边组成的有向无环图，可以最多删除k条边让他的拓扑序最大。输出最大的拓扑序。

思路：在以前的topsort中是入读为零的点入队列，这里有k次机会可以删除边，那么我就把所有入度<=k的点全入队列，用优先队列维护最大的点序列号，去掉点最大序列号的所有入边，将它加入到拓扑序中，这样贪心是最优的。

13278437

2015-03-29 09:39:39

Accepted

5195

374MS

8344K

2822 B

C++

wust_lyf

13278469

2015-03-29 09:42:51

Accepted

5195

1996MS

10928K

2822 B

G++

wust_lyf

G++和C++差别如此大




代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 200005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

struct Node
{
    int x;
    int id;
    friend bool operator<(const Node a,const Node b)
    {
        return a.id<b.id;
    }
};

int n,m,k;
vector <int> edge[maxn];
priority_queue<Node> Q;

int inDegree[maxn];
int ans[maxn];
int vis[maxn];

int main()
{
    int i;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        for (i=0;i<=n+2;i++)
        {
            edge[i].clear();
            inDegree[i]=0;
            vis[i]=0;  //不在队列中
            ans[i]=0;
        }
        for (i=0;i<m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            inDegree[b]++;
            edge[a].push_back(b);
        }

        while(!Q.empty())
            Q.pop();
        Node node;

        for (i=1;i<=n;i++)
            if (inDegree[i]<=k)
            {
                node.id=i;
                node.x=inDegree[i];
                Q.push(node);
                vis[i]=1;//在队列中
            }
        int t=0;

        while (!Q.empty())
        {
            Node now=Q.top();
            Q.pop();
            if (inDegree[now.id]<=k)
                k-=inDegree[now.id];
            else
            {
                vis[now.id]=0;
                continue;
            }
            vis[now.id]=2;//已经输出
            ans[t++]=now.id;
            for (i=0;i<edge[now.id].size();i++)
            {
                int v=edge[now.id][i];
                inDegree[v]--;
                node.id=v;
                node.x=inDegree[v];
                if (inDegree[v]<=k && vis[v]==0)
                {
                    Q.push(node);
                    vis[v]=1;
                }
            }
        }

        pf("%d",ans[0]);
        for (int i=1;i<t;i++)
            pf(" %d",ans[i]);
        pf("\n");
    }
    return 0;
}
/*
6 6 2
5 6
1 2
4 5
2 4
3 4
2 3
3 2 0
1 2
1 3
*/