POJ - 2492 A Bug's Life
Background
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.
Problem
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.
Problem
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.
Input
The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.
Output
The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.
Sample Input
2 3 3 1 2 2 3 1 3 4 2 1 2 3 4
Sample Output
Scenario #1: Suspicious bugs found! Scenario #2: No suspicious bugs found!
Hint
Huge input,scanf is recommended.
1 /* 2 #include <stdio.h> 3 #include <iostream> 4 using namespace std; 5 //存储的是其父亲的下表 6 int bugs[2010]; 7 int relation[2010];//1:相同性别 0:不同性别 8 //初始化 9 void init(int len) 10 { 11 for(int i = 0;i <= len; i++) 12 { 13 bugs[i] = i; 14 relation[i] = 1; 15 } 16 } 17 //找到根 18 int find(int bug) 19 { 20 if(bugs[bug]==bug)return bug; 21 int tem = bugs[bug]; 22 bugs[bug] = find(bugs[bug]);//递归更新域,返回最终的父亲节点,把所有的孩子都更新了 23 //注意这里,求当前位置和父亲的关系,记录之前父亲的位置为tem,然后因为是递归, 24 //此时的relation[tem]已经在递归中更新过了,也就是孩子和父亲的关系+父亲和爷爷的关系+1然后模2就得到 25 //孩子和爷爷的关系,这里用0和1表示,0表示不同性别,1表示相同性别 26 relation[bug] = (relation[bug]+relation[tem]+1)%2; 27 return bugs[bug]; 28 } 29 30 void union_set(int a,int b,int x,int y) 31 { 32 //合并,让前边的集合的根指向后边集合的根,成为一个集合 33 bugs[x]=y; 34 //更新前边集合根和新的集合根之间的关系, 35 //注意这里,relation[a]+relation[x]与relation[b] 36 //相对于新的父节点必须相差1个等级,因为他们不是gay 37 relation[x] = (relation[b]-relation[a])%2; 38 } 39 40 int main() 41 { 42 int S; 43 int n,inter; 44 int bug1,bug2,parent1,parent2; 45 bool flag;//false:无同性恋,true:有同性恋 46 scanf("%d",&S); 47 for(int i=1; i<=S;i++) 48 { 49 scanf("%d%d",&n,&inter); 50 flag = false; 51 init(n);//初始化,使其父节点为自己 52 for(int j = 1; j <= inter; j++) 53 { 54 scanf("%d%d",&bug1,&bug2); 55 if(flag)continue; 56 parent1 = find(bug1); 57 parent2 = find(bug2); 58 if(parent1==parent2) 59 { 60 if(relation[bug1]==relation[bug2])//同性 61 flag = true; 62 } 63 union_set(bug1,bug2,parent1,parent2); 64 } 65 if(flag) 66 printf("Scenario #%d:\nSuspicious bugs found!\n",i); 67 else 68 printf("Scenario #%d:\nNo suspicious bugs found!\n",i); 69 printf("\n"); 70 } 71 return 0; 72 } 73 74 wa 了不知为什么,, 75 */ 76 #include <cstdio> 77 #include <iostream> 78 #include <cstring> 79 80 using namespace std; 81 82 const int MAX = 2005; 83 84 bool flag; 85 int fa, fb, n; 86 int father[MAX], rela[MAX]; 87 88 int find(int x) 89 { 90 if(father[x] == -1) 91 return x; 92 else 93 { 94 int tmp = father[x]; 95 father[x] = find(father[x]); 96 97 rela[x] = (rela[x]+rela[tmp])%2; 98 return father[x]; 99 } 100 } 101 102 int main() 103 { 104 int T, m, a, b; 105 scanf("%d", &T); 106 for(int k = 1; k <= T; ++k) 107 { 108 flag = true; 109 scanf("%d%d", &n, &m); 110 { 111 memset(father, -1, (n+1)*sizeof(int)); 112 memset(rela, 0, (n+1)*sizeof(int)); 113 while(m--) 114 { 115 scanf("%d%d", &a, &b); 116 if(!flag) 117 continue; 118 fa = find(a); 119 fb = find(b); 120 121 if(fa != fb) 122 { 123 father[fa] = fb; 124 rela[fa] = (rela[a] + rela[b] + 1)%2; 125 } 126 else 127 { 128 if(rela[a] == rela[b]) 129 flag = false; 130 } 131 132 } 133 134 printf("Scenario #%d:\n", k); 135 if(flag) 136 printf("No suspicious bugs found!\n\n"); 137 else 138 printf("Suspicious bugs found!\n\n"); 139 } 140 } 141 return 0; 142 }