标签:include efi 连通 style main with const template double
题目链接:https://vjudge.net/problem/POJ-2236
题目大意:有一个计算机网络的所有线路都坏了,网络中有n台计算机,现在你可以做两种操作,修理(O)和检测两台计算机是否连通(S),只有修理好的计算机才能连通。连通有个规则,两台计算机的距离不能超过给定的最大距离D(一开始会给你n台计算机的坐标)。检测的时候输出两台计算机是否能连通。
只要每次修好一台计算机的同时把周围距离小于等于d的计算机加入集合就行
#include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl ‘\n‘ #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define zero(a) memset(a, 0, sizeof(a)) #define INF(a) fill(a, a+maxn, INF); #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; typedef pair<double, int> P2; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1000000007LL; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const double EULC = 0.5772156649015328; const int NIL = -1; template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == ‘-‘)f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } const int maxn = 1e3+10; int p[maxn], vis[maxn]; P cp[maxn]; int __find(int root) { int son = root, temp; while(root != p[root]) root = p[root]; while(son != root) { temp = p[son]; p[son] = root; son = temp; } return root; } void merge(int son1, int son2) { p[__find(son1)] = p[__find(son2)]; } int main(void) { IOS; int n, d; while(cin >> n >> d) { zero(vis); for (int i = 0; i<n; ++i) p[i] = i; for (int i = 0; i<n; ++i) cin >> cp[i].first >> cp[i].second; string s; while(cin >> s) { if (s[0] == ‘O‘) { int pos; cin >> pos; vis[pos-1] = true; for (int i = 0; i<n; ++i) if (i != pos-1 && vis[i] && (cp[i].first-cp[pos-1].first)*(cp[i].first-cp[pos-1].first) + (cp[i].second-cp[pos-1].second)*(cp[i].second-cp[pos-1].second) <= d*d) merge(i, pos-1); } else if (s[0] == ‘S‘) { int a, b; cin >> a >> b; if (__find(a-1) == __find(b-1)) printf("SUCCESS\n"); else printf("FAIL\n"); } } } return 0; }
POJ - 2236 - Wireless Network(并查集)
标签:include efi 连通 style main with const template double
原文地址:https://www.cnblogs.com/shuitiangong/p/12384737.html