hoj 2218 Median Weight Bead // poj 1975 Median Weight Bead 有向图的传递闭包问题
2012-03-29 15:53
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/*
有向图的传递闭包问题,只要顶点的入度或出度大于n/2,即可判断它不符合中间点,
即答案加一,而求它的入度或出度的话,可以用有向图的传递闭包floyd算法来做,
时间复杂度为O(n^3)
*/
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define X 100
int map[X][X],n,m;
void floyd()
{
for(int k=1;k<=n;k++) //floyd算法求传递闭包
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
map[i][j] = map[i][j]||(map[i][k]&&map[k][j]);
}
int main()
{
freopen("sum.in","r",stdin);
freopen("sum.out","w",stdout);
int t,x,y;
cin>>t;
while(t--)
{
memset(map,0,sizeof(map));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
map[x][y] = 1; //表示从顶点x到y处在有向边
}
floyd();
int ans = 0;
for(int i=1;i<=n;i++)
{
int sum1 = 0;
int sum2 = 0;
for(int j=1;j<=n;j++) //求顶点的入度或出度
{
if(i==j)
continue;
if(map[i][j]) //如果为出度
sum1++;
if(map[j][i]) //为入度
sum2++;
if(sum1>n/2||sum2>n/2)
{
ans++;
break;
}
}
}
cout<<ans<<endl;
}
return 0;
}
有向图的传递闭包问题,只要顶点的入度或出度大于n/2,即可判断它不符合中间点,
即答案加一,而求它的入度或出度的话,可以用有向图的传递闭包floyd算法来做,
时间复杂度为O(n^3)
*/
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define X 100
int map[X][X],n,m;
void floyd()
{
for(int k=1;k<=n;k++) //floyd算法求传递闭包
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
map[i][j] = map[i][j]||(map[i][k]&&map[k][j]);
}
int main()
{
freopen("sum.in","r",stdin);
freopen("sum.out","w",stdout);
int t,x,y;
cin>>t;
while(t--)
{
memset(map,0,sizeof(map));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
map[x][y] = 1; //表示从顶点x到y处在有向边
}
floyd();
int ans = 0;
for(int i=1;i<=n;i++)
{
int sum1 = 0;
int sum2 = 0;
for(int j=1;j<=n;j++) //求顶点的入度或出度
{
if(i==j)
continue;
if(map[i][j]) //如果为出度
sum1++;
if(map[j][i]) //为入度
sum2++;
if(sum1>n/2||sum2>n/2)
{
ans++;
break;
}
}
}
cout<<ans<<endl;
}
return 0;
}
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