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 51 21 31 41 51 615 21 21 32

• 样例输出

• 42

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;}