Posted on 2010-08-10 15:18
MiYu 閱讀(516)
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ACM ( 并查集 )
MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)載自 ______________白白の屋
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1213
題目描述:
How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2337 Accepted Submission(s): 1033
Problem Description
Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.
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.
Input
The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines 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.
Output
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
Sample Input
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
題目分析:
并查集中的超級(jí)水題, 只要判斷集合的個(gè)數(shù)就可以了....................
代碼如下:
MiYu原創(chuàng), 轉(zhuǎn)帖請(qǐng)注明 : 轉(zhuǎn)載自 ______________白白の屋
#include <iostream>
using namespace std;
typedef struct {
int parent;
int cnt;
}Tset;
typedef struct treeUFS{
public:
treeUFS(int n = 0):N(n+1) { set = new Tset[N]; for ( int i = 0; i != N; ++ i)
set[i].parent = i,set[i].cnt = 1;
}
~treeUFS() { delete [] set; };
int find ( int x ){ int r = x; while ( set[r].parent != r ) //循環(huán)結(jié)束，則找到根節(jié)點(diǎn)
r = set[r].parent; int i = x;
//本循環(huán)修改查找路徑中所有節(jié)點(diǎn)
while ( i != r) {
int j = set[i].parent; set[i].parent = r; i = j;
}
return r;
}
void init () { for ( int i = 0; i != N; ++ i) set[i].parent = i,set[i].cnt = 1; }
int getSetCount ( int x ){ return set[ find(x) ].cnt; }
void Merge( int x,int y ){ x = find ( x ); y = find ( y );
if ( x == y ) return;
if ( set[x].cnt > set[y].cnt ){
set[y].parent = x;
set[x].cnt += set[y].cnt;
}
else{ set[x].parent = y;
set[y].cnt += set[x].cnt;
}
}
private:
Tset *set;
int N;
}treeUFS;
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
int N,M;
scanf ( "%d%d",&N,&M );
treeUFS UFS ( N );
for ( int i = 1; i <= M; ++ i )
{
int a,b;
scanf ( "%d%d",&a,&b );
UFS.Merge ( a,b );
}
int nCount = 0;
for ( int i = 1; i <= N; ++ i )
{
if ( UFS.find (i) == i )
{
nCount ++;
}
}
printf ( "%d\n",nCount );
}
return 0;
}