每个物品有三个参量，其实等价于两个，因为总和确定。
于是问题变成二维平面上一堆点，求最小的b的子集形成的凸包，包含这些点a。
用Floyd绕一圈即可。
1 /************************************************************** 2 Problem: 1027 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1136 ms 7 Memory:1820 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 #define P Point 15 using namespace std; 16 typedef double lf; 17 const int N = 505; 18 const int inf = 1e9; 19 const lf eps = 1e-8; 20 21 struct Point { 22 lf x, y; 23 P() {} 24 P(lf _x, lf _y) : x(_x), y(_y) {} 25 26 inline bool operator != (const P p) const { 27 return fabs(x - p.x) > eps || fabs(y - p.y) > eps; 28 } 29 inline P operator - (P p) { 30 return P(x - p.x, y - p.y); 31 } 32 inline lf operator * (P p) { 33 return x * p.y - y * p.x; 34 } 35 36 inline void read() { 37 scanf("%lf%lf%*lf", &x, &y); 38 } 39 } a[N], b[N]; 40 41 int n, m; 42 int dis[N][N]; 43 44 bool in_one_line(P x, P y) { 45 int i; 46 if (x.x > y.x) swap(x, y); 47 for (i = 1; i <= m; ++i) 48 if (b[i].x < x.x || b[i].x > y.x) return 0; 49 if (x.y > y.y) swap(x, y); 50 for (i = 1; i <= m; ++i) 51 if (b[i].y < x.y || b[i].y > y.y) return 0; 52 return 1; 53 } 54 55 int check(P x, P y) { 56 int i, cnt1, cnt2; 57 lf tmp; 58 for (i = 1, cnt1 = cnt2 = 0; i <= m; ++i) { 59 tmp = (y - x) * (b[i] - x); 60 if (tmp > eps) ++cnt1; 61 if (tmp < -eps) ++cnt2; 62 if (cnt1 && cnt2) return 0; 63 } 64 if (!cnt1 && !cnt2 && in_one_line(x, y)) { 65 puts("2"); 66 return -1; 67 } 68 return cnt1 ? 1 : cnt2 ? 2 : 3; 69 } 70 71 void Floyd() { 72 int ans = inf, i, j, k; 73 for (k = 1; k <= n; ++k) 74 for (i = 1; i <= n; ++i) 75 for (j = 1; j <= n; ++j) 76 dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]); 77 for (i = 1; i <= n; ++i) 78 ans = min(ans, dis[i][i]); 79 if (ans == inf || ans <= 2) puts("-1"); 80 else printf("%d ", ans); 81 } 82 83 void solve() { 84 int i, j, f; 85 for (i = 1; i <= n; ++i) 86 for (j = 1; j <= n; ++j) 87 dis[i][j] = inf; 88 for (i = 1; i <= n; ++i) 89 for (j = i + 1; j <= n; ++j) { 90 f = check(a[i], a[j]); 91 if (f == -1) return; 92 if (f == 1) dis[i][j] = 1; 93 else if (f == 2) dis[j][i] = 1; 94 else if (f == 3) dis[i][j] = dis[j][i] = 1; 95 } 96 Floyd(); 97 } 98 99 bool special_judge() { 100 int i; 101 for (i = 1; i <= n; ++i) 102 if (a[1] != a[i]) return 0; 103 for (i = 1; i <= m; ++i) 104 if (a[1] != b[i]) return 0; 105 puts("1"); 106 return 1; 107 } 108 109 int main() { 110 int i; 111 scanf("%d%d", &n, &m); 112 for (i = 1; i <= n; ++i) 113 a[i].read(); 114 for (i = 1; i <= m; ++i) 115 b[i].read(); 116 if (special_judge()) return 0; 117 solve(); 118 return 0; 119 }