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这次的题让我对选择不同数据结构所产生的结果惊呆了,一开始用的是结构来存储集合,课件上有现成的,而且我也是实在不太会,150ms的时间限制过不去,不得已,看到这题刚好可以用数组,结果7ms最多,有意思!什么是时间复杂度,终于有了一次感性的认识。这题还有个check的要求,我用了一个链表来存储,感觉挺麻烦的,不知是否有更好的方法,题设要求及代码实现如下
1 /* 2 Name: 3 Copyright: 4 Author: 5 Date: 06/04/15 09:46 6 Description: 7 We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other? 8 9 Input Specification: 10 11 Each input file contains one test case. For each test case, the first line contains N (2<=N<=104), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format: 12 13 I c1 c2 14 where I stands for inputting a connection between c1 and c2; or 15 16 C c1 c2 17 where C stands for checking if it is possible to transfer files between c1 and c2; or 18 19 S 20 where S stands for stopping this case. 21 22 Output Specification: 23 24 For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network. 25 26 Sample Input 1: 27 5 28 C 3 2 29 I 3 2 30 C 1 5 31 I 4 5 32 I 2 4 33 C 3 5 34 S 35 Sample Output 1: 36 no 37 no 38 yes 39 There are 2 components. 40 Sample Input 2: 41 5 42 C 3 2 43 I 3 2 44 C 1 5 45 I 4 5 46 I 2 4 47 C 3 5 48 I 1 3 49 C 1 5 50 S 51 Sample Output 2: 52 no 53 no 54 yes 55 yes 56 The network is connected. 57 */ 58 #include <stdio.h> 59 #include <stdlib.h> 60 #include <stdbool.h> 61 62 typedef struct Flag 63 { 64 bool flag; 65 struct Flag * next; 66 }Flag, * pFlag; 67 68 int Find(int S[], int X); 69 void Union(int S[], int X1, int X2); 70 71 int MaxSize; 72 73 int main() 74 { 75 // freopen("in.txt", "r", stdin); // for test 76 int i, c1, c2, k; 77 char ch; 78 79 scanf("%d\n", &MaxSize); 80 81 int S[MaxSize]; 82 for(i = 0; i < MaxSize; i++) 83 S[i] = -1; 84 85 pFlag head = (pFlag)malloc(sizeof(Flag)); 86 pFlag p, tmp; 87 88 p = head; 89 scanf("%c", &ch); 90 while(ch != ‘S‘) 91 { 92 if(ch == ‘C‘) 93 { 94 pFlag f = (pFlag)malloc(sizeof(Flag)); 95 f->next = NULL; 96 scanf("%d%d\n", &c1, &c2); 97 if(Find(S, c1) == Find(S, c2)) 98 f->flag = true; 99 else 100 f->flag = false; 101 p->next = f; 102 p = p->next; 103 } 104 else 105 { 106 scanf("%d%d", &c1, &c2); 107 Union(S, c1, c2); 108 } 109 scanf("%c", &ch); 110 } 111 112 p = head->next; 113 tmp = head; 114 free(tmp); 115 while(p) 116 { 117 if(p->flag) 118 printf("yes\n"); 119 else 120 printf("no\n"); 121 tmp = p; 122 p = p->next; 123 free(tmp); 124 } 125 k = 0; 126 do 127 { 128 if(S[--MaxSize] < 0) 129 k++; 130 }while(MaxSize); 131 if(k > 1) 132 printf("There are %d components.\n", k); 133 else 134 printf("The network is connected.\n"); 135 // fclose(stdin); // for test 136 return 0; 137 } 138 139 int Find(int S[], int X) 140 { 141 X--; 142 for(; S[X] >= 0; X = S[X]); 143 144 return X; 145 } 146 147 void Union(int S[], int X1, int X2) 148 { 149 int Root1, Root2; 150 151 Root1 = Find(S, X1); 152 Root2 = Find(S, X2); 153 if(Root1 != Root2) 154 { 155 if(S[Root1] <= S[Root2]) 156 { 157 S[Root1] += S[Root2]; 158 S[Root2] = Root1; 159 } 160 else 161 { 162 S[Root2] += S[Root1]; 163 S[Root1] = Root2; 164 } 165 } 166 }
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原文地址:http://www.cnblogs.com/qingkai/p/4395994.html