HDU 4582 DFS spanning tree（DFS+贪心）（2013ACM-ICPC杭州赛区全国邀请赛）

Problem Description

Consider a Depth-First-Search(DFS) spanning tree T of a undirected connected graph G, we define a T-Simple Circle as a path v ₁, v ₂, ..., v _k (v ₁ = v _k) in G that contains at most one edge which not belongs to the DFS spanning tree T.
Given a graph G, we process DFS on G starting from vertex 1 and get a DFS spanning tree T, then you should choose some edges of G so that all T-Simple Circles contains at least one edge that you choose.
Please minimize the number of edges you choose.

 

Input

There are at most 100 test cases.
For each case, the first line contains two integers n and m denoting the number of vertices and edges. The vertexes are numbered from 1 to n.
The following m lines describe the graph. Each line contains two integers x _i and y _i, denoting an edge between vertex x _i and y _i(x _i ≠ y _i).
Note that the first n-1 edges of input construct a DFS spanning tree T which is generated by DFS from vertex 1.
Input ends with n = 0 and m = 0
(1 <= n <= 2000, 1 <= m <= 20000, 1 <= x _i, y _i <= n)

 

Output

For each case, output the number of minimal edges that you choose.

 

题目大意&思路： http://blog.csdn.net/sd_invol/article/details/9963741

PS：有个大于号写反了居然过了样例……样例不要这么坑爹好吗……

 

代码（812MS）：

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <algorithm>

 4 #include <queue>

 5 #include <cstring>

 6 using namespace std;

 7 

 8 const int MAXN = 2010;

 9 const int MAXE = 20010;

10 

11 int dep[MAXN];

12 

13 struct Edge {

14     int x, y, id;

15     void read(int i) {

16         id = i;

17         scanf("%d%d", &x, &y);

18     }

19     void adjust() {

20         if(dep[x] > dep[y]) swap(x, y);

21     }

22     bool operator < (const Edge &rhs) const {

23         return dep[x] > dep[rhs.x];

24     }

25 } e[MAXE];

26 

27 int n, m;

28 int head[MAXN], fa[MAXN];

29 int to[MAXN * 2], next[MAXN * 2];

30 int ecnt;

31 

32 void init() {

33     memset(head, 0, sizeof(head));

34     ecnt = 1;

35 }

36 

37 void add_edge2(int u, int v) {

38     to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;

39     to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;

40 }

41 

42 void bfs() {

43     memset(dep, -1, sizeof(dep));

44     queue<int> que; que.push(1);

45     dep[1] = 0;

46     while(!que.empty()) {

47         int u = que.front(); que.pop();

48         for(int p = head[u]; p; p = next[p]) {

49             int &v = to[p];

50             if(dep[v] == -1) {

51                 fa[v] = u;

52                 dep[v] = dep[u] + 1;

53                 que.push(v);

54             }

55         }

56     }

57 }

58 

59 bool vis[MAXN];

60 

61 bool check(Edge &p) {

62     int now = p.y;

63     while(fa[now] != p.x) {

64         if(vis[now]) break;

65         now = fa[now];

66     }

67     if(!vis[now]) {

68         vis[now] = true;

69         return false;

70     }

71     else return true;

72 }

73 

74 int main() {

75     while(scanf("%d%d", &n, &m) != EOF) {

76         if(n == 0 && m == 0) break;

77         for(int i = 1; i <= m; ++i) e[i].read(i);

78         init();

79         for(int i = 1; i < n; ++i) add_edge2(e[i].x, e[i].y);

80         bfs();

81         for(int i = n; i <= m; ++i) e[i].adjust();

82         sort(e + 1, e + m + 1);

83         memset(vis, 0, sizeof(vis));

84         int ans = 0;

85         for(int i = 1; i <= m; ++i)

86             if(e[i].id >= n && !check(e[i])) ++ans;

87         printf("%d\n", ans);

88     }

89 }

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