输入
输出
示例输入
4 1 0 1 0 2 0 3 0 4 O 1 O 2 O 4 S 1 4 O 3 S 1 4
示例输出
FAIL SUCCESS
思路:
对于O操作,每修理一个电脑就检测该电脑与所有被检测过的电脑之间能否通信,可以就把两个电脑加入同一个集合中
对于S操作,判断两台电脑是否在都修理过且在同一个集合中,满足就输出SUCCESS,否则输出FAIL
代码:
#include <iostream> #include <algorithm> #include <string.h> #include <cstdio> #include <string> #include <cmath> #include <vector> #include <stack> #include <queue> #include <stack> #include <list> #include <map> #include <set> //#include <unordered_map> #define Fbo friend bool operator < (node a, node b) #define mem(a, b) memset(a, b, sizeof(a)) #define FOR(a, b, c) for (int a = b; a <= c; a++) #define RFOR(a, b, c) for (int a = b; a >= c; a--) #define off ios::sync_with_stdio(0) #define sc(a) scanf("%d",&a) bool check1(int a) { return (a & (a - 1)) == 0 ? true : false; } using namespace std; typedef pair<int, int> pii; typedef long long ll; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; const int Maxn = 1e5 + 5; const double pi = acos(-1.0); const double eps = 1e-8; int a[Maxn],n,d; int b[Maxn], cnt=0;//被修理的电脑编号 bool vis[Maxn]; struct node { int x, y; }point[Maxn]; void init() { for (int i = 1; i < Maxn; i++) a[i] = i; } int find(int x) { return x == a[x] ? x : find(a[x]); } void unite(int x, int y) { x = find(x); y = find(y); if (x != y) { a[x] = a[y]; } } bool check(int a, int b) { double dis = sqrt((double)((point[a].x - point[b].x) * (point[a].x - point[b].x)) + ((point[a].y - point[b].y) * (point[a].y - point[b].y))); //cout << dis << "---" << d << endl; return dis <= d; } int main() { off; mem(vis, 0); cin >> n >> d; init(); for (int i = 1; i <= n; i++) { cin >> point[i].x >> point[i].y; } char ch; while (cin >> ch) { if (ch == ‘O‘) { int m; cin >> m; vis[m] = 1; for (int i = 0; i < cnt; i++) { if (check(b[i], m)) { unite(b[i], m); } } b[cnt++] = m; } else { int l, r; cin >> l >> r; if (vis[l] && vis[r] && find(l) == find(r)) { cout << "SUCCESS" << endl; } else cout << "FAIL" << endl; } } return 0; }
(并查集模板题)Wireless Network POJ - 2236
原文:https://www.cnblogs.com/AlexLINS/p/12656000.html