Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 8797 | Accepted: 4116 |
Description
possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure
occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.
Input
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;
Output
Sample Input
5 5 1 2 3 4 0 6 2 1 3 5 4 6 2 0 0
Sample Output
1 2
Hint
//題意,給一個圖,求圖中割頂的個數~~~
#include "stdio.h" //poj 1144 割頂的判斷模板
#include "vector"
#include "string.h"
using namespace std;
#define N 105
vector<int> a[N];
int root;
int low[N],dfn[N];
bool visit[N],mark[N];
int MIN(int x,int y) { return x<y?x:y; }
void DFS(int x,int times)
{
int i;
int count = 0;
visit[x] = true;
low[x] = dfn[x] = times;
for(i=0; i<a[x].size(); ++i)
{
int y = a[x][i];
if(!visit[y])
{
count++;
DFS(y,times+1);
low[x] = MIN(low[x],low[y]);
if(x==root && count==2) mark[root] = true; //如果x爲根節點,並且子樹有兩個或兩個以上,則爲割頂!
if(x!=root && low[y]>=dfn[x]) mark[x] = true; //如果x的子樹中沒有節點連接x的祖先(有的話low[x]會變小),則爲割頂!
}
else
low[x] = MIN(low[x],dfn[y]);//更新low[x]的值!
}
}
int main()
{
int n;
int i,j;
char ch;
while(scanf("%d",&n),n!=0)
{
int x,y;
for(i=0; i<N; ++i)
a[i].clear();
while(scanf("%d",&x),x!=0)
{
while(scanf("%c",&ch) && ch==' ')
{
scanf("%d",&y);
a[y].push_back(x);
a[x].push_back(y);
}
}
root = 1;
int times = 1;
memset(mark,false,sizeof(mark)); //標記點是否是割頂
memset(visit,false,sizeof(visit)); //標記點是否已訪問
DFS(root,times);
int ans = 0;
for(i=1; i<=n; ++i)
{
if(mark[i]) ans++; //統計割頂個數
}
printf("%d\n",ans);
}
return 0;
}