题目描述
该题的目的是要你统计图的连通分支数。
输入
每个输入文件包含若干行,每行两个整数i,j,表示节点i和j之间存在一条边。
输出
输出每个图的联通分支数。
样例输入
1 4
4 3
5 5
样例输出
2
分析: 由于题目没给出范围,只能大概猜测点的范围,数组至少要开到10的6次方。这里我是用并查集计算的。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;
int findFather(int father[],int x)
{
if(father[x]==-1)return -1;
int a=x;
while(x!=father[x])x=father[x];
while(a!=father[a])
{
int z=a;
a=father[a],father[z]=x;
}
return x;
}
void Union(int a,int b,int father[])
{
int fa=findFather(father,a),fb=findFather(father,b);
if(fa!=fb)
father[fa]=fb;
return;
}
int father[1000010];
bool isroot[1000010];
int main(void)
{
#ifdef test
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
clock_t start=clock();
#endif //test
for(int i=0;i<1000010;++i)
father[i]=-1;
int a,b;
while(~scanf("%d%d",&a,&b))
{
if(father[a]==-1)father[a]=a;
if(father[b]==-1)father[b]=b;
Union(a,b,father);
}
for(int i=0;i<1000010;++i)
{
int index=findFather(father,i);
if(index!=-1)isroot[index]=1;
}
int ans=0;
for(int i=0;i<1000010;++i)
ans+=isroot[i];
printf("%d\n",ans);
#ifdef test
clockid_t end=clock();
double endtime=(double)(end-start)/CLOCKS_PER_SEC;
printf("\n\n\n\n\n");
cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位
cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位
#endif //test
return 0;
}