Hdoj 5195 Dzy Loves topological sorting "topology" + "line segment tree"

Dzy Loves Topological sorting

Time limit:4000/2000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 922 Accepted Submission (s): 269

Problem Description
A topological sort or topological ordering of a directed graph is a linear ordering of it vertices such that for every Di Rected Edge (u→v) from vertex U-vertex v, U comes before v in the ordering.
Now, the 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 a 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

Hint
Case 1.
Erase The Edge (2->3), (4->5).
and the lexicographically largest topological ordering is (5,3,1,2,4).

Test instructions: give you n-side, delete not more than the K-bar, so that the output of the dictionary order the largest!!
Strategy: We always look for a larger number than the current K output is good,
Need to understand: 1, each minus one entry is subtracted from an edge.
2: After finding a point, be sure to change the entry of the corresponding point to the maximum value, in case there is a possibility of being found later.
Code:

#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <algorithm>Const intM =1e5+5;Const intINF =0x3f3f3f3f;using namespace STD;intc[m<<2], in[m]; vector<int >M[M]; vector<int >AnsintN, MM, K;voidUpdateintPintXintLintRintPOS) {if(L = = r) {C[pos] = x;return; }intMid = (l+r) >>1;if(P <= mid) Update (p, X, L, Mid, pos<<1);//left and right are the corresponding points of the representation.     ElseUpdate (P, X, mid+1, R, pos<<1|1); C[pos] = min (c[pos<<1], c[pos<<1|1]);}intQueryintLintRintPOS) {if(L = = r)returnLintMid = (l+r) >>1;if(c[pos<<1|1] <= k)returnQuery (mid+1, R, pos<<1|1);//Always try to choose the bigger point    returnQuery (L, mid,pos<<1);}voidTopo () { for(inti =0; I < n; + + i) {inttemp = Query (1N1); K-= in[temp];//means to remove several pointsAns.push_back (temp); Update (temp, INF,1N1);//To update when foundIn[temp] = INF;//must become positive infinity         for(inti =0; I < m[temp].size (); + + i) {intv = m[temp][i];             --IN[V]; Update (V, In[v],1N1); }    }}intMain () { while(scanf("%d%d%d", &n, &mm, &k) = =3){ for(inti =0; I <= N; + + i) {m[i].clear (); In[i] =0; }intU, v; for(inti =0; I < mm; + + i) {scanf("%d%d", &u, &v);            ++IN[V];        M[u].push_back (v); } for(inti =1; I <= N; + + i) {update (I, In[i],1N1);        } ans.clear (); Topo ();printf("%d", ans[0]); for(inti =1; I < n; + + i)printf("%d", Ans[i]);printf("\ n"); }return 0;}

Hdoj 5195 Dzy Loves topological sorting "topology" + "line segment tree"