The Cow Prom
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1132 | Accepted: 713 |
Description
The N (2 <= N <= 10,000) cows are so
excited: it‘s prom night! They are dressed in their finest gowns, complete with
corsages and new shoes. They know that tonight they will each try to perform the
Round Dance.
Only cows can perform the Round Dance which requires a
set of ropes and a circular stock tank. To begin, the cows line up around a
circular stock tank and number themselves in clockwise order consecutively from
1..N. Each cow faces the tank so she can see the other
dancers.
They then acquire a total of M (2 <= M <= 50,000)
ropes all of which are distributed to the cows who hold them in their hooves.
Each cow hopes to be given one or more ropes to hold in both her left and right
hooves; some cows might be disappointed.
For the Round Dance to
succeed for any given cow (say, Bessie), the ropes that she holds must be
configured just right. To know if Bessie‘s dance is successful, one must examine
the set of cows holding the other ends of her ropes (if she has any), along with
the cows holding the other ends of any ropes they hold, etc. When Bessie dances
clockwise around the tank, she must instantly pull all the other cows in her
group around clockwise, too. Likewise,
if she dances the other way, she
must instantly pull the entire group counterclockwise (anti-clockwise in British
English).
Of course, if the ropes are not properly distributed then
a set of cows might not form a proper dance group and thus can not succeed at
the Round Dance. One way this happens is when only one rope connects two cows.
One cow could pull the other in one direction, but could not pull the other
direction (since pushing ropes is well-known to be fruitless). Note that the
cows must Dance in lock-step: a dangling cow (perhaps with just one rope) that
is eventually pulled along disqualifies a group from properly performing the
Round Dance since she is not immediately pulled into lockstep with the
rest.
Given the ropes and their distribution to cows, how many
groups of cows can properly perform the Round Dance? Note that a set of ropes
and cows might wrap many times around the stock tank.
Input
Line 1: Two space-separated integers: N and
M
Lines 2..M+1: Each line contains two space-separated integers A
and B that describe a rope from cow A to cow B in the clockwise direction.
Output
Line 1: A single line with a single integer that
is the number of groups successfully dancing the Round Dance.
Sample Input
5 4
2 4
3 5
1 2
4 1
Sample Output
1
Hint
Explanation of the sample:
ASCII art
for Round Dancing is challenging. Nevertheless, here is a representation of the
cows around the stock tank:
_1___
/****
5 /****** 2
/ /**TANK**|
\ \********/
\ \******/ 3
\ 4____/ /
\_______/
Cows 1, 2, and 4 are properly connected and form a complete Round Dance group.
Cows 3 and 5 don‘t have the second rope they‘d need to be able to pull both
ways, thus they can not properly perform the Round Dance.
Source
强连通分量
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <stack>
6
7 using namespace std;
8
9 const int MAX_N = 10005;
10 const int edge = 50005;
11 int N,M;
12 int first[MAX_N],Next[edge],v[edge];
13 int low[MAX_N],pre[MAX_N],cmp[MAX_N];
14 int num[MAX_N];
15 int dfs_clock,scc_cnt;
16 stack<int> S;
17
18
19 void add_edge(int id,int u) {
20 int e = first[u];
21 Next[id] = e;
22 first[u] = id;
23 }
24
25 void dfs(int u) {
26 pre[u] = low[u] = ++dfs_clock;
27 S.push(u);
28 for(int e = first[u]; e != -1; e = Next[e]) {
29 if(!pre[ v[e] ]) {
30 dfs( v[e] );
31 low[u] = min(low[u],low[ v[e] ]);
32 } else if(!cmp[ v[e] ]) {
33 low[u] = min(low[u],pre[ v[e] ]);
34 }
35 }
36
37 if(low[u] == pre[u]) {
38 ++scc_cnt;
39 for(;;) {
40 int x = S.top(); S.pop();
41 cmp[x] = scc_cnt;
42 num[scc_cnt]++;
43 if(x == u) break;
44 }
45 }
46 }
47
48 void scc() {
49 dfs_clock = scc_cnt = 0;
50 memset(cmp,0,sizeof(cmp));
51 memset(pre,0,sizeof(pre));
52
53 for(int i = 1; i <= N; ++i) if(!pre[i]) dfs(i);
54 }
55
56 void solve() {
57 int ans = 0;
58 scc();
59 for(int i = 1; i <= scc_cnt; ++i) {
60 if(num[i] >= 2) ++ans;
61 }
62
63 printf("%d\n",ans);
64 }
65 int main()
66 {
67 、、freopen("sw.in","r",stdin);
68 scanf("%d%d",&N,&M);
69 for(int i = 1; i <= N; ++i) first[i] = -1;
70 for(int i = 0; i < M; ++i) {
71 int u;
72 scanf("%d%d",&u,&v[i]);
73 add_edge(i,u);
74 }
75
76 solve();
77 return 0;
78 }