题目链接:click here
思路:挺抽象的一道题,他没说有多少个节点,只给出道路以及其相连的点,那所有的节点就是已知道路对应的节点总和。这里用set存储最合适,因为要去重。剩下就是标准的并查集了。另外这题没给出数据范围,那么久老规矩100w。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <climits>
using namespace std;
const int MAXN = 1000005;
const int INF = INT_MAX;
int father[MAXN];
int height[MAXN];
int Find(int x){
if(x != father[x]) father[x] = Find(father[x]);
return father[x];
}
void Union(int x, int y){
x = Find(x);
y = Find(y);
if(x != y){
if(height[x] < height[y]) father[x] = y;
else if(height[y] < height[x]) father[y] = x;
else{
father[y] = x;
height[x]++;
}
}
}
void Initial(){
for(int i = 0; i < MAXN; i++){
father[i] = i;
height[i] = 0;
}
}
int main(){
// freopen("in.txt", "r", stdin);
int i, j;
Initial();
set<int> myset;
while(~scanf("%d %d", &i, &j)){
Union(i, j);
myset.insert(i);
myset.insert(j);
}
int ans = 0;
set<int>::iterator it;
for(it = myset.begin(); it != myset.end(); it++){
int current = *it;
if(current == father[current]) ans++;
}
printf("%d\n", ans);
return 0;
}