The Cow Prom
| Time Limit: 1000MS |
Memory Limit: 65536K |
|
| Total Submissions: 2541 |
Accepted: 1503 |
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 42 43 51 24 1Sample Output
1Hint
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
算法分析：
题意：
给你 N 个点和 M 条边，求大小 >1 的强连通分量。
分析：
用 Tarjan求强连通分量就可以了，求解的同时维护每个强连通分量的大小，最后输出大小 >1 的强连通分量的个数即可。
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const int maxn=10000+10;
int n,m;
vector<int> G[maxn];
stack<int> S;
int dfs_clock, scc_cnt;
int pre[maxn],low[maxn],sccno[maxn];
int num[maxn];//记录每个分量的节点数
bool out0[maxn];//标记该分量是否出度为0
void dfs(int u)
{
pre[u]=low[u]=++dfs_clock;
S.push(u);
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(!pre[v])
{
dfs(v);
low[u]=min(low[u],low[v]);
}
else if(!sccno[v])
low[u]=min(low[u],pre[v]);
}
if(low[u]==pre[u])
{
scc_cnt++;
num[scc_cnt]=0;
while(true)
{
int x=S.top(); S.pop();
sccno[x]=scc_cnt;
num[scc_cnt]++;
if(x==u) break;
}
}
}
void find_scc(int n)
{
dfs_clock=scc_cnt=0;
memset(pre,0,sizeof(pre));
memset(sccno,0,sizeof(sccno));
for(int i=1;i<=n;i++)
if(!pre[i]) dfs(i);
}
int main()
{
while(scanf("%d",&n)==1&&n)
{
scanf("%d",&m);
for(int i=1;i<=n;i++) G[i].clear();
while(m--)
{
int u,v;
scanf("%d%d",&u,&v);
G[u].push_back(v);
}
find_scc(n);
int tot=0;
for(int i=1;i<=scc_cnt;i++)
if(num[i]>1) tot++;
cout<<tot<<endl;
}
return 0;
}