http://poj.org/problem?id=2236
并查集
预处理每两台之间的距离 这样不用再后面重复计算
用一个数组dist[i][j]存储
修好一台 就枚举N 合并集合
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 struct Computer 7 { 8 int x, y; 9 bool repair; 10 }computer[1007]; 11 12 bool near[1007][1007]; //预处理 两台电脑之间 距离的关系 13 int par[1007]; 14 15 int get_dist(int a, int b) 16 { 17 if (a == b) return 0; 18 return (computer[a].x - computer[b].x)*(computer[a].x - computer[b].x)+(computer[a].y - computer[b].y)*(computer[a].y - computer[b].y); 19 } 20 int find(int x) 21 { 22 if (par[x] == x) return x; 23 else return par[x] = find(par[x]); 24 } 25 26 void unite(int x, int y) 27 { 28 int px = find(x), py = find(y); 29 if (px == py) return ; 30 else par[py] = px; 31 } 32 33 bool same(int x, int y) 34 { 35 int px = find(x), py = find(y); 36 return px == py; 37 } 38 39 40 int main() 41 { 42 freopen("in.txt", "r", stdin); 43 int N; 44 int D; 45 scanf("%d%d", &N, &D); 46 D*=D;//变成D^2 47 memset(near, 0, sizeof(near)); 48 for (int i = 1; i <= N; i++) 49 { 50 int x, y; 51 par[i] = i; 52 scanf("%d%d",&x, &y); 53 computer[i].y = y; 54 computer[i].repair = false; 55 } 56 for (int i = 1; i <= N; i++) 57 { 58 for (int j = 1; j <= N; j++) 59 { 60 if (get_dist(i, j) <= D ) 61 { 62 near[i][j] = 1; 63 near[j][i] = 1; 64 } 65 } 66 } 67 char ch; 68 getchar(); 69 while (~scanf("%c", &ch)) 70 { 71 int c1, c2; 72 if (ch == 'O') 73 { 74 scanf("%d", &c1); 75 computer[c1].repair = true; 76 for (int i = 1; i<= N; i++) 77 { 78 if (near[i][c1] && computer[i].repair) unite(c1, i); 79 } 80 getchar(); 81 82 } 83 else 84 { 85 scanf("%d%d", &c1, &c2); 86 getchar(); 87 if (computer[c1].repair && computer[c2].repair && same(c1, c2)) printf("SUCCESS ");//必须要都修好了 88 else printf("FAIL "); 89 } 90 } 91 return 0; 92 }