poj——3660(图论之传递闭包(floyd))
2013-08-31 19:44
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题目地址:http://poj.org/problem?id=3660
注:floyd的另个运用。
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
#define MAX(a,b) a>b?a:b
#define MIN(a,b) a>b?b:a
#define N 105
int a
;
int main()
{
int i,j,k,m,n,num1,num2;
while(cin>>n>>m)
{
memset(a,0,sizeof(a));
while(m--)
{
cin>>num1>>num2;
a[num1][num2]=1;
}
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
for(k=1;k<=n;k++)
{
if(a[j][i]&&a[i][k]) a[j][k]=1;
}
int ans=0;
for(i=1;i<=n;i++)
{
int temp=0;
for(j=1;j<=n;j++)
temp+=(a[i][j]+a[j][i]);
if(temp==n-1) ans++;
}
cout<<ans<<endl;
}
return 0;
}
注:floyd的另个运用。
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
#define MAX(a,b) a>b?a:b
#define MIN(a,b) a>b?b:a
#define N 105
int a
;
int main()
{
int i,j,k,m,n,num1,num2;
while(cin>>n>>m)
{
memset(a,0,sizeof(a));
while(m--)
{
cin>>num1>>num2;
a[num1][num2]=1;
}
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
for(k=1;k<=n;k++)
{
if(a[j][i]&&a[i][k]) a[j][k]=1;
}
int ans=0;
for(i=1;i<=n;i++)
{
int temp=0;
for(j=1;j<=n;j++)
temp+=(a[i][j]+a[j][i]);
if(temp==n-1) ans++;
}
cout<<ans<<endl;
}
return 0;
}
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