【并查集专题】【HDU】
PS:做到第四题才发现 2,3题的路径压缩等于没写
How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14542 Accepted Submission(s): 7132
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
2
5 3
1 2
2 3
4 5 5 1
2 5
2
4
裸的并查集没什么好说的
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
int N,M;
int Set[1001];
int ans;
void YCL()
{
ans=0;
for(int i=0;i<=N;i++)
Set[i]=i;
}
int find(int x)
{
if(x!=Set[x])
Set[x]=find(Set[x]);
return Set[x];
}
int Union(int a,int b)
{
int a1=find(a);
int b1=find(b);
if(a1!=b1)
{
Set[a1]=b1;
return 1;
}
else return 0;
}
void input()
{
cin>>N>>M;
int a,b;
YCL();
for(int i=1;i<=M;i++)
{
cin>>a>>b;
if(Union(a,b))
ans++;
}
}
int main()
{
int T;
cin>>T;
while(T--)
{
input();
cout<<N-ans<<endl;
}
return 0;
}
小希的迷宫
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27792 Accepted Submission(s): 8589

整个文件以两个-1结尾。
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
Yes
Yes
No
几点注意:
1.不止要判环,还要判断是否是一棵树,而不是多棵树
2.注意0 0 //,1 1 0 0 这样的数据
3.编号非连续的,所以开个数组存编号
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
int father[100001];
int A[200001];
int Max=-1,tot=0;
int find(int x)
{
if(x!=father[x])
father[x]=find(father[x]);
return father[x];
}
int Union(int a,int b)
{
int a1=find(a);
int a2=find(b);
if(a1==a2) return 0;
else father[a1]=a2;
return 1;
}
void init()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
}
void YCL()
{
for(int i=1;i<=100000;i++)
{
father[i]=i;
}
}
int main()
{
// init();
int a,b,OK=1;
while(1)
{
YCL();
OK=1;Max=-1;tot=0;
while(cin>>a>>b)
{
if(a==-1&&b==-1) return 0;
else if(a==0&&b==0) {break;}
else if(OK)
if(Union(a,b)==0) OK=0;
A[++tot]=a;
A[++tot]=b;
}
if(OK==1) {
for(int i=1;i<=tot;i++)
if(find(father[A[i]])!=find(father[A[1]])) OK=0;
if(OK==1) printf("Yes\n");
else printf("No\n");
}
else cout<<"No"<<endl;
}
return 0;
}
Is It A Tree?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15095 Accepted Submission(s): 3342
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.
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.
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
Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.
注意 这和上面的题有点不一样 是一个有向图
所以要判断入度
否则会出现
判断成树的情况
代码如下:
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
int father[100001];
int A[200001];
int T[200001];
int Max=-1,tot=0;
int find(int x)
{
if(x!=father[x])
father[x]=find(father[x]);
return father[x];
}
int Union(int a,int b)
{
int a1=find(a);
int a2=find(b);
if(a1==a2) return 0;
else father[a1]=a2;
return 1;
}
void init()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
}
void YCL()
{
for(int i=1;i<=tot;i++)
{
father[A[i]]=A[i];
T[A[i]]=0;
}
}
int main()
{
// init();
int a,b,OK=1;
int kk=0;
while(1)
{
kk++;
OK=1;Max=-1;tot=0;
while(cin>>a>>b)
{
if(a<0||b<0) return 0;
else if(a==0&&b==0) {break;}
A[++tot]=a;
A[++tot]=b;
}
YCL();
for(int i=1;i<=tot/2;i++)
{
if(OK)
{
if(Union(A[2*(i-1)+1],A[2*i])==0) OK=0;
T[A[2*i]]++;
}
}
if(OK==1) {
for(int i=1;i<=tot;i++)
{
if(find(father[A[i]])!=find(father[A[1]])) OK=0;
if(T[A[i]]>1) OK=0;
}
if(OK==1) printf("Case %d is a tree.\n",kk);
else printf("Case %d is not a tree.\n",kk);
}
else printf("Case %d is not a tree.\n",kk);
}
return 0;
}
More is better
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 15588 Accepted Submission(s): 5737
Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are
still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
4
2HintA and B are friends(direct or indirect), B and C are friends(direct or indirect),
then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.
MAP离散化+路径压缩。。 还一直以为是MAP导致TLE 结果发现是2,3题的路径压缩写错了。。
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#include<map>
#define oo 0x13131313
using namespace std;
int N;
int father[200000];
int A[200000];
int B[200000];
int num[200000];
map<int,int> Hash;
int tot=0;
int ans=1;
void input()
{
Hash.clear();
tot=0;ans=1;
int a,b;
for(int i=1;i<=N;i++)
{
scanf("%d%d",&a,&b);
B[(2*i-1)]=A[(2*i-1)]=a;
B[(2*i)]=A[(2*i)]=b;
}
}
void CSH()
{
B[0]=-1;
sort(B+1,B+2*N+1);
for(int i=1;i<=2*N;i++)
{
if(B[i]!=B[i-1])
Hash[B[i]]=++tot;
}
for(int i=1;i<=tot;i++)
{
father[i]=i;
num[i]=1;
}
}
int find(int x)
{
if(x!=father[x])
father[x]=find(father[x]);
return father[x];
}
int Union(int a,int b)
{
int a1=find(a);
int b1=find(b);
if(a1==b1) return 0;
else
{
father[a1]=b1;
num[b1]+=num[a1];
if(num[b1]>ans) ans=num[b1];
}
return 1;
}
void solve()
{
for(int i=1;i<=N;i++)
{
Union(Hash[A[(2*i-1)]],Hash[A[(2*i)]]);
}
}
void init()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
}
int main()
{
// init();
while(scanf("%d",&N)!=EOF)
{
input();
CSH();
solve();
printf("%d\n",ans);
}
return 0;
}
Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14672 Accepted Submission(s): 5572
C such that there is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
i and village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
3
0 990 692
990 0 179
692 179 0
1
1 2
179
给定N条建好路径的最小生成树
1.建好的路径就是权值为0即可,或直接克鲁斯卡尔
附代码:
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
struct edge
{
int s,t,w;
};
int father[101];
edge A[20111];
int map[101][101];
int N;
int ans;
int tot=0;
int Q;
int k;
int cmp(const void *i,const void *j)
{
edge *ii=(edge *)i;edge *jj=(edge *)j;
return ii->w-jj->w;
}
void YCL()
{
for(int i=1;i<=101;i++)
{
father[i]=i;
}
}
int find(int x)
{
if(x!=father[x])
father[x]=find(father[x]);
return father[x];
}
int Union(int a,int b)
{
int a1=find(a);
int b1=find(b);
if(a1==b1) return 0;
else
{
father[b1]=a1;
}
return 1;
}
void input()
{
YCL();
int temp,a,b;
ans=0;
tot=0;
k=0;
for(int i=1;i<=N;i++)
for(int j=1;j<=N;j++)
{
scanf("%d",&map[i][j]);
if(i<j)
{
tot++;
A[tot].s=i;A[tot].t=j;A[tot].w=map[i][j];
}
}
cin>>Q;
for(int i=1;i<=Q;i++)
{
scanf("%d%d",&a,&b);
if(Union(a,b))
{
k++;
}
}
}
void solve()
{
int j=1;
qsort(A+1,tot,sizeof(A[1]),cmp);
for(int i=1;i<=N-1-k;i++)
{
while(Union(A[j].s,A[j].t)==0&&j<=tot)
j++;
ans+=A[j].w;
}
}
int main()
{
while(cin>>N)
{
input();
solve();
cout<<ans<<endl;
}
}
后面还有几道生成树水题就不上了。。
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