題目: UVa - 10245 - The Closest Pair Problem
題目說明 給一些在二維平面上的點,求最近的兩點距離為何。
Input: 每組測資第一個整數 N
,表示有幾個點 ( 若 N
為 0 表結束 ),後面 N
行每行有兩個整數,表點的座標。
Output: 輸出最近的兩點距離為何,輸出至小數點後四位,若超過 10000 則輸出 "INFINITY"
。
解題思路 Closet Pair Problem,可用 Sweep line 或 Divide and Conquer。
參考解法 Sweep line:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 #include <bits/stdc++.h> using namespace std ;#define FOR(i, a, b) for (int i = a; i < b; ++i) #define CLR(c) memset(c, 0, sizeof(c)) static auto __ = []{ ios_base::sync_with_stdio(0 ); cin .tie(0 ); cout .tie(0 ); return 0 ; }(); const int INF = (int )1e9 ;const int MXN = 10000 ;struct Point { double x; double y; }; int N;Point P[MXN]; double dist (const Point& l, const Point& r) { return sqrt (pow (r.x - l.x, 2 ) + pow (r.y - l.y, 2 )); } void read () { FOR(i, 0 , N) cin >> P[i].x >> P[i].y; } void solve () { sort(P, P + N, [](Point& l, Point& r) { return l.x < r.x; }); double d = 1e4 ; FOR(i, 0 , N) FOR(j, i + 1 , N) { if (P[i].x + d < P[j].x) break ; d = min(dist(P[i], P[j]), d); } if (d == 1e4 ) cout << "INFINITY\n" ; else cout << setprecision(4 ) << fixed << d << '\n' ; } int main () { while (cin >> N && N) { read(); solve(); } }
Divide and Conquer:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 #include <bits/stdc++.h> using namespace std ;#define FOR(i, a, b) for (int i = a; i < b; ++i) #define CLR(c) memset(c, 0, sizeof(c)) static auto __ = []{ ios_base::sync_with_stdio(0 ); cin .tie(0 ); cout .tie(0 ); return 0 ; }(); const int INF = (int )1e9 ;const int MXN = 10000 ;struct Point { double x; double y; }; Point P[MXN]; Point tmp[MXN]; int N;double dist (const Point& l, const Point& r) { return sqrt (pow (l.x - r.x, 2 ) + pow (l.y - r.y, 2 )); } bool cmp1 (const Point& l, const Point& r) { return l.x < r.x; } bool cmp2 (const Point& l, const Point& r) { return l.y < r.y; } void read () { FOR(i, 0 , N) cin >> P[i].x >> P[i].y; } double DnC (int L, int R) { if (L >= R) return INF; auto M = (L + R) / 2 ; auto d = min(DnC(L, M), DnC(M + 1 , R)); int idx = 0 ; for (int i = M; i >= L && P[M].x - P[i].x < d; --i) tmp[idx++] = P[i]; for (int i = M + 1 ; i <= R && P[i].x - P[M].x < d; ++i) tmp[idx++] = P[i]; sort(tmp, tmp + idx, cmp2); FOR(i, 0 , idx) FOR(j, 1 , 4 && i + j < idx) d = min(dist(tmp[i], tmp[i + j]), d); return d; } void solve () { sort(P, P + N, cmp1); auto ret = DnC(0 , N - 1 ); if (ret >= 1e4 ) cout << "INFINITY\n" ; else cout << setprecision(4 ) << fixed << ret << '\n' ; } int main () { while (cin >> N && N) { read(); solve(); } }
參考資料 UVa 10245 - The Closest Pair Problem | NaiveRed&’s Blog UVa 10245 - The Closest Pair Problem_小白菜又菜-CSDN博客