HDU 1272
I - 小希的迷宫
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Appoint description:
Description
上次Gardon的迷宫城堡小希玩了很久（见Problem B），现在她也想设计一个迷宫让Gardon来走。但是她设计迷宫的思路不一样，首先她认为所有的通道都应该是双向连通的，就是说如果有一个通道连通了房 间A和B，那么既可以通过它从房间A走到房间B，也可以通过它从房间B走到房间A，为了提高难度，小希希望任意两个房间有且仅有一条路径可以相通（除非走 了回头路）。小希现在把她的设计图给你，让你帮忙判断她的设计图是否符合她的设计思路。比如下面的例子，前两个是符合条件的，但是最后一个却有两种方法从 5到达8。
Input
输入包含多组数据，每组数据是一个以0 0结尾的整数对列表，表示了一条通道连接的两个房间的编号。房间的编号至少为1，且不超过100000。每两组数据之间有一个空行。
整个文件以两个-1结尾。
整个文件以两个-1结尾。
Output
对于输入的每一组数据，输出仅包括一行。如果该迷宫符合小希的思路，那么输出"Yes"，否则输出"No"。
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
Yes Yes No
本体就是判断整个图有没有环并且时不时联通的
1判断有没有环直接用在加边的时候判断根节点是否相同就可以了，如果想同，那么肯定形成了环
2判断连通性是在所有的都合并好了以后，暴力循环判断，如果所有点的根节点都一样，那么肯定是联通的，如果发现有不一样的了，那么就不会连在一起
3有一个特殊的地方就是如果只是0 0的话，那么我们默认这个是联通的输出yes
代码
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
int father[maxn];
bool vis[maxn];
void init(){
for(int i=;i<maxn;i++){
father[i]=i;
vis[i]=false;
}
}
int get_father(int x){
if(father[x]!=x)
father[x]=get_father(father[x]);
return father[x];
}
int main(){
int u,v;
bool flag=true;
init();
while(scanf("%d%d",&u,&v)!=EOF){
if(u==-&&v==-)
break;
if(u==&&v==){ printf("Yes\n");
continue;
}
vis[u]=true;
vis[v]=true;
father[u]=v;
while(scanf("%d%d",&u,&v)!=EOF){
if(u==&&v==)
break;
vis[u]=true;
vis[v]=true;
int t1=get_father(u);
int t2=get_father(v);
if(t1!=t2)
father[t1]=t2;
else
flag=false;
}
int ans;
for(int i=;i<maxn;i++){
if(vis[i]){
ans=get_father(i);
break;
}
}
for(int i=;i<maxn;i++){
if(vis[i]){
if(get_father(i)!=ans){
flag=false;
break;
} }
}
if(flag)
printf("Yes\n");
else
printf("No\n");
flag=true;
init(); }
return ; }
H - Is It A Tree?
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Appoint description:
Description
A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.
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
The input will consist of a sequence of descriptions (test cases)
followed by a pair of negative integers. Each test case will consist of a
sequence of edge descriptions followed by a pair of zeroes Each edge
description will consist of a pair of integers; the first integer
identifies the node from which the edge begins, and the second integer
identifies the node to which the edge is directed. Node numbers will
always be greater than zero.
followed by a pair of negative integers. Each test case will consist of a
sequence of edge descriptions followed by a pair of zeroes Each edge
description will consist of a pair of integers; the first integer
identifies the node from which the edge begins, and the second integer
identifies the node to which the edge is directed. Node numbers will
always be greater than zero.
Output
For each test case display the line "Case k is a tree." or the line
"Case k is not a tree.", where k corresponds to the test case number
(they are sequentially numbered starting with 1).
"Case k is not a tree.", where k corresponds to the test case number
(they are sequentially numbered starting with 1).
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.
本题和上题几乎一样，只不过这道题判断的时不时一个树。
这里我们需要注意三点
1本题给的虽然是有向图，但是并不影响树的构建，当成无向图就可以了，用并查集合并的时候，代码和上道题是一模一样的 2如果输入的数据只有一组，那么一个点是孤立的节点，他是构不成一个树的
3如果只是输入0 0 ，那么这是一个空树，但是也是一个树，所以我们仍需要默认其为一个树
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
int father[maxn];
bool vis[maxn];
void init(){
for(int i=;i<maxn;i++){
father[i]=i;
vis[i]=false;
}
}
int get_father(int x){
if(father[x]!=x)
father[x]=get_father(father[x]);
return father[x];
}
int main(){
int u,v;
bool flag=true;
init();
int cnt=;
while(scanf("%d%d",&u,&v)!=EOF){
if(u==-&&v==-)
break;
if(u==&&v==){ printf("Case %d is a tree.\n",cnt++);
continue;
}
vis[u]=true;
vis[v]=true;
father[u]=v;
while(scanf("%d%d",&u,&v)!=EOF){
if(u==&&v==)
break;
vis[u]=true;
vis[v]=true;
int t1=get_father(u);
int t2=get_father(v);
if(t1!=t2)
father[t1]=t2;
else
flag=false;
}
int ans;
int sum=;
for(int i=;i<maxn;i++){
if(vis[i]){
ans=get_father(i);
break;
}
}
for(int i=;i<maxn;i++){
if(vis[i]){
sum++;
if(get_father(i)!=ans){
flag=false;
break;
} }
}
if(flag&&sum!=)
printf("Case %d is a tree.\n",cnt++);
else
printf("Case %d is not a tree.\n",cnt++);
flag=true;
init(); }
return ; }