A Bug's Life |
| Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 364 Accepted Submission(s): 140 |
| Problem Description 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. |
| 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 23 31 22 31 34 21 23 4 |
| Sample Output Scenario #1:Suspicious bugs found!Scenario #2:No suspicious bugs found! |
分析：
（1）题意很简单，就是检查一堆数据中是否有同性恋，找出主要矛盾是如果1喜欢2,2喜欢3，而1又喜欢3，则矛盾。找出出现类的的矛盾
（2）这里用bugs数组存储其父亲节点的下标，用relation数组存储该虫子与根节点之间的关系。1为同性，0为异性。
（3）并查集详见注释
#include//存储的是其父亲的下表int bugs[2010];int relation[2010];//1:相同性别 0：不同性别//初始化void init(int len){ for(int i = 0;i <= len; i++) { bugs[i] = i; relation[i] = 1; }}//找到根int find(int bug){ if(bugs[bug]==bug)return bug; int tem = bugs[bug]; 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 //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， //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 //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 relation[bug] = (relation[bug]+relation[tem]+1)%2; return bugs[bug];}void union_set(int a,int b,int x,int y){ //合并，让前边的集合的根指向后边集合的根，成为一个集合 bugs[x]=y; //更新前边集合根和新的集合根之间的关系， //注意这里，relation[a]+relation[x]与relation[b] //相对于新的父节点必须相差1个等级，因为他们不是gay relation[x] = (relation[b]-relation[a])%2;}int main(){ int S; int n,inter; int bug1,bug2,parent1,parent2; bool flag;//false:无同性恋，true：有同性恋 scanf("%d",&S); for(int i=1; i<=S;i++) { scanf("%d%d",&n,&inter); flag = false; init(n);//初始化，使其父节点为自己 for(int j = 1; j <= inter; j++) { scanf("%d%d",&bug1,&bug2); if(flag)continue; parent1 = find(bug1); parent2 = find(bug2); if(parent1==parent2) { if(relation[bug1]==relation[bug2])//同性 flag = true; } union_set(bug1,bug2,parent1,parent2); } if(flag) printf("Scenario #%d:\nSuspicious bugs found!\n",i); else printf("Scenario #%d:\nNo suspicious bugs found!\n",i); printf("\n"); } return 0;}