hdu 5195(拓扑排序+线段树优化)

DZY Loves Topological Sorting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1064    Accepted Submission(s): 327

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 mostk 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 21 24 52 43 42 33 2 01 21 3

 

Sample Output

5 3 1 2 41 3 2
Hint
Case 1.Erase the edge (2->3),(4->5).And the lexicographically largest topological ordering is (5,3,1,2,4).
 
解题思路：
         因为我们要求最后的拓扑序列字典序最大，所以一定要贪心地将标号越大的点越早入队。我们定义点i的入度为di。假设当前还能删去k条边，         那么我们一定会把当前还没入队的di≤k的最大的i找出来，把它的di条入边都删掉，然后加入拓扑序列。可以证明，这一定是最优的。具体实         现可以用线段树维护每个位置的di，在线段树上二分可以找到当前还没入队的di≤k的最大的i。于是时间复杂度就是O((n+m)logn).
#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<set>using namespace std;const int maxn = 100005;const int inf = 0x3f3f3f3f;struct Seg {int l,r,Min; //Min表示在区间[l,r]上的最小入度}tree[maxn<<2];struct Edge {int to,next;}edge[maxn];int n,m,k,cnt,head[maxn];int ind[maxn],res[maxn];void addedge(int u,int v) {edge[cnt].to = v;edge[cnt].next = head[u];head[u] = cnt++;}void build(int rt,int l,int r) {tree[rt].l = l, tree[rt].r = r;if(tree[rt].l == tree[rt].r) {tree[rt].Min = ind[l];return;}int mid = (l + r) >> 1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);tree[rt].Min = min(tree[rt<<1].Min,tree[rt<<1|1].Min);}void update(int rt,int pos) {if(tree[rt].l == tree[rt].r) {tree[rt].Min--;return;}int mid = (tree[rt].l + tree[rt].r) >> 1;if(pos <= mid)update(rt<<1,pos);else update(rt<<1|1,pos);tree[rt].Min = min(tree[rt<<1].Min,tree[rt<<1|1].Min);}int query(int rt,int k) {if(tree[rt].l == tree[rt].r) {tree[rt].Min = inf;return tree[rt].l;}int ans;if(tree[rt<<1|1].Min <= k) ans = query(rt<<1|1,k);else ans = query(rt<<1,k);tree[rt].Min = min(tree[rt<<1].Min,tree[rt<<1|1].Min);return ans;}int main() {int u,v;while(scanf("%d%d%d",&n,&m,&k)!=EOF) {cnt = 0;memset(head,-1,sizeof(head));memset(ind,0,sizeof(ind));for(int i = 1; i <= m; i++){scanf("%d%d",&u,&v);addedge(u,v);ind[v]++;}build(1,1,n);for(int i = 1; i <= n; i++) {int u = query(1,k);printf(i==n?"%d\n":"%d ",u);k -= ind[u];ind[u] = 0;for(int j = head[u]; j != -1; j = edge[j].next) {int v = edge[j].to;if(ind[v] > 0) {ind[v]--;update(1,v);}}}}return 0;}