POJ 1308 Is It A Tree? (并查集)
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24237 | Accepted: 8311 |
Description
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not. 
In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.
Input
Output
Sample Input
6 8 5 3 5 2 6 4
5 6 0 0 8 1 7 3 6 2 8 9 7 5
7 4 7 8 7 6 0 0 3 8 6 8 6 4
5 3 5 6 5 2 0 0
-1 -1
Sample Output
Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree. 和HDU 1272一模一样,除了改下输出其他什么都不用改。
#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
using namespace std; const int SIZE = ;
int FATHER[SIZE],RANK[SIZE];
vector<int> S; void ini(int);
int find_father(int);
bool unite(int,int);
bool same(int,int);
int main(void)
{
int a,b,count = ;
bool flag; while(scanf("%d%d",&a,&b) && (a != - && b != -))
{
flag = true;
if(a == && b == )
{
printf("Case %d %s\n",++ count,flag ? "is a tree." : "is not a tree.");
continue;
}
S.clear();
S.push_back(a);
S.push_back(b);
ini(SIZE - );
flag = unite(a,b);
while(scanf("%d%d",&a,&b) && (a || b))
{
flag &= unite(a,b);
S.push_back(a);
S.push_back(b);
}
for(int i = ;i < S.size();i ++)
flag &= same(S[],S[i]);
printf("Case %d %s\n",++ count,flag ? "is a tree." : "is not a tree.");
} return ;
} void ini(int n)
{
for(int i = ;i <= n;i ++)
{
FATHER[i] = i;
RANK[i] = ;
}
} int find_father(int n)
{
if(n == FATHER[n])
return n;
return FATHER[n] = find_father(FATHER[n]);
} bool unite(int x,int y)
{
x = find_father(x);
y = find_father(y); if(x == y)
return false;
if(RANK[x] < RANK[y])
FATHER[x] = y;
else
{
FATHER[y] = x;
if(RANK[x] == RANK[y])
RANK[x] ++;
}
return true;
} bool same(int x,int y)
{
return find_father(x) == find_father(y);
}
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