題目鏈接:http://poj.org/problem?id=2186
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 38414 | Accepted: 15657 |
Description
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow.
Input
* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.
Output
Sample Input
3 3 1 2 2 1 2 3
Sample Output
1
思路:這其實是一道Tarjin的模板題(水題)。。。
不過之前看到這道題是在白書上,後來查了一下,那個算法叫Kosaraju算法,今天又補了個Tarjin的做法,兩者各有千秋吧。
順便又搞了個Tarjin的模板,美滋滋 (o゚▽゚)o
代碼 :
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
using namespace std;
const int MAXN = 10010;
const int MAXM = 50010;
int du[MAXN];
int Low[MAXN], DFN[MAXN], Stack[MAXN], Belong[MAXN];
int Index, top;
int scc;
bool Instack[MAXN];
int num[MAXN];
vector<int> g[MAXN];
void Tarjan(int u)
{
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
int len=g[u].size();
for (int i = 0; i < len; i++)
{
v=g[u][i];
if (!DFN[v])
{
Tarjan(v);
if (Low[u] > Low[v])
{
Low[u] = Low[v];
}
}
else if (Instack[v] && Low[u] > DFN[v])
{
Low[u] = DFN[v];
}
}
if (Low[u] == DFN[u])
{
scc++;
do
{
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc;
num[scc]++;
}
while (v != u);
}
return ;
}
void solve(int N)
{
memset(DFN, 0, sizeof(DFN));
memset(Instack, false, sizeof(Instack));
Index = scc = top = 0;
for (int i = 1; i <= N; i++)
{
if (!DFN[i])
{
Tarjan(i);
}
}
return ;
}
int main()
{
int m,n;
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
g[x].push_back(y);
}
solve(n);
for(int i=1;i<=n;i++)
{
int len=g[i].size();
for(int j=0;j<len;j++)
{
int cur=g[i][j];
if(Belong[i]!=Belong[cur])
{
du[Belong[i]]++;
}
}
}
int sum=0;
int ind;
for(int i=1;i<=scc;i++)
{
if(du[i]==0)
{
sum++;
ind=i;
}
}
if(sum!=1)
printf("0\n");
else
{
printf("%d\n",num[ind]);
}
return 0;
}