题意：判断一个给定的图，没有环，而且存在一个链，图上的所有点或者在这条链上或者在其的邻居
题解：
1.判断环：
对于无向图：如果 点 < 边 + 1，则存在环；
然后使用并查集进一步判断环的存在
2.判断是否存在链
首先统计各个点的度，然后对于度为1的点，将其相连的边删掉，再统计新图的度，这时新图应该剩下一条链，
也就是说，新图的不存在大于2个度为1的点，而且这个点在旧图的度是大于1的。
代码：
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
using namespace std;
const int N = 105;
vector<int> graphs[N];
int deg[N];
int degOld[N];
int n,e;
int cnt;
class UnionSet
{
private:
int parent[N];
int rank[N];
int size;
public:
UnionSet()
{
size = 0;
}
UnionSet(int _size)
{
init(_size);
}
~UnionSet()
{
}
void init(int _size)
{
size = _size;
for (int i = 0; i < size; ++i)
{
parent[i] = -1;
rank[i] = 1;
}
}
void adjust(int _root,int _x)
{
int i = _x;
int j ;
while(parent[i] >= 0)
{
j = parent[i];
parent[i] = _root;
i = j;
}
}
int getRoot(int _x)
{
int r = _x;
while(parent[r] >= 0)
{
r= parent[r];
}
//adjust
adjust(r,_x);
return r;
}
void join(int _r1,int _r2)
{
if (_r1 == _r2)
{
return ;
}
int root1 = getRoot(_r1);
int root2 = getRoot(_r2);
if (root1 == root2)
{
return ;
}
if (rank[root1] > rank[root2])
{
parent[root2] = root1;
rank[root1] += rank[root2];
}
else
{
parent[root1] = root2;
rank[root2] += rank[root1];
}
}
int getRank(int _x)
{
return rank[_x];
}
};
static UnionSet uSet;
void Test()
{
memset(deg,0,sizeof(deg));
for (int i = 0; i < N; ++i)
{
graphs[i].clear();
}
int a,b;
uSet.init(n+1);
for (int i = 0; i < e; ++i)
{
scanf("%d %d",&a,&b);
deg[a]++;
deg[b]++;
graphs[a].push_back(b);
graphs[b].push_back(a);
uSet.join(a,b);
}
if (n < e + 1)
{
printf("Graph %d is not a caterpillar.\n", cnt);
return ;
}
for (int i = 2; i <= n; ++i)
{
if (uSet.getRoot(1) != uSet.getRoot(i))
{
printf("Graph %d is not a caterpillar.\n", cnt);
return ;
}
}
for (int i = 1; i <= n; ++i)
{
degOld[i] = deg[i];
}
for (int i = 1; i <= n; ++i)
{
if (degOld[i] == 1)
{
for (int j = 0; j < graphs[i].size(); ++j)
{
deg[graphs[i][j]]--;
}
}
}
int tmp = 0;
for (int i = 1; i <= n; ++i)
{
if(degOld[i] > 1 && deg[i] == 1)
++tmp;
}
if(tmp > 2)
printf("Graph %d is not a caterpillar.\n", cnt);
else
printf("Graph %d is a caterpillar.\n", cnt);
}
int main()
{
//freopen("data.txt","r",stdin);
cnt = 0;
while(scanf("%d",&n) != EOF)
{
if(n == 0)
break;
scanf("%d",&e);
++cnt;
Test();
}
return 0;
}
posted on 2011-11-17 10:50
bennycen 阅读(5977)
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算法题解