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题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1213
这题粗略一看好熟悉,原来和 POJ 2524 Ubiquitous Religions 一模一样,就是输出时不用输出Case情况计数。Orz
// Author: Tanky Woo
// HDOJ 1213
// Accepted 1213 0MS 200K 1004 B C++ Tanky Woo
#include
using namespace std;
#define MAX 100001
// father[x]表示x的父节点
int father[MAX];
// rank[x]表示x的秩
int rank[MAX];
// 初始化
void Make_Set(int n)
{
for(int i=1; i<=n; ++i)
{
father[i] = i;
rank[i] = 0;
}
}
// 查找
int Find_Set(int x)
{
if(x != father[x])
return Find_Set(father[x]);
return x;
}
// 合并
void Union(int x, int y)
{
x = Find_Set(x);
y = Find_Set(y);
if(x == y) // x,y在同一个集合
return;
if(rank[x] > rank[y])
father[y] = x;
else if(rank[x] < rank[y])
father[x] = y;
else
{
rank[y]++;
father[x] = y;
}
}
int main()
{
freopen("input.txt", "r", stdin);
int nCases;
int a, b;
scanf("%d", &nCases;);
while(nCases--)
{
int N, M;
scanf("%d %d", &N;, &M;);
Make_Set(N);
for(int i=0; i<M; ++i)
{
scanf("%d %d", &a;, &b;);
a = Find_Set(a);
b = Find_Set(b);
if(a != b)
{
Union(a, b);
N--;
}
}
printf("%d\n", N);
}
return 0;
}