NOJ——1274The battle of Red Cliff（并查集按秩合并）

• [1274] The battle of Red Cliff

• 时间限制: 1000 ms　内存限制: 65535 K
• 问题描述

• Zero loves palying Sanguo game very much, The battle of Red Cliff is one of scenes. Now he will put a fire to one of ships. Some ships are connected, some are not.Now, zero puts his fire, please tell him how many ships will be left.

  

• 输入
• Input until EOF. 
  There are two positive integers N(2<N<100) and M(0<M<100), N means the number of ships, M means the number of command. 
  Then M lines follows. Each line has two positive integers x and y means he will connect these two ships by a rope. 
  Last line will include a number of ship's that zero will burn. 
• 输出
• The ship's number is from 1 N, every ship which is connected a burning ship will also be burned.Now, please tell zero how many ships will be left. 
• 样例输入

• 10 5
  1 2
  1 3
  1 4
  1 5
  1 6
  1
  5 2
  1 2
  1 3
  2

• 样例输出

• 4
  2

WA数次，后来换了个思路，直接求被烧掉的船所在集合的元素个数即可。答案就是总船数减掉集合元素个数。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
#define MM(a) memset(a,0,sizeof(a))
int pos[1100];
int rank[1100];
inline int find(int x)
{
	if(x!=pos[x])
		return pos[x]=find(pos[x]);
	return pos[x];
}
inline void joint(int a,int b)
{
	int aa=find(a),bb=find(b);
	if(aa!=bb)
	{
		if(rank[aa]<rank[bb])
		{
			pos[aa]=bb;
			rank[bb]+=rank[aa];
		}
		else
		{
			pos[bb]=aa;
			rank[aa]+=rank[bb];
		}
	}
}
int main(void)
{
	int m,n,i,ans,x,y,t;
	while (~scanf("%d%d",&n,&m))
	{
		MM(pos);MM(rank);
		for (i=0; i<1100; i++)
		{
			pos[i]=i;
			rank[i]=1;
		}
		for (i=1; i<=m; i++)
		{
			scanf("%d%d",&x,&y);
			joint(x,y);
		}
		scanf("%d",&t);
		ans=0;
		t=find(t);
		printf("%d\n",n-rank[t]);
	}
	return 0;
}