poj2492_A Bug's Life_并查集

A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 34947   Accepted: 11459 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 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. Source TUD Programming Contest 2005, Darmstadt, Germany

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 1 #include <iostream>
 2 #include <cstdio>
 3 #define MAX_N 150000+5
 4 
 5 using namespace std;
 6 
 7 int par[MAX_N];//父节点
 8 int depth[MAX_N];//深度
 9 
10 void init(int n){
11    for(int i=0;i<=n;i++){
12        par[i]=i;
13        depth[i]=1;
14    }
15 }
16 int find_father(int t){
17    if(t==par[t]){
18        return t;
19    }else{
20        return par[t]=find_father(par[t]);
21        //实现了路径压缩
22    }
23 }
24 void unite(int t1,int t2){
25    int f1=find_father(t1);
26    int f2=find_father(t2);
27    if(f1==f2){
28        return ;
29    }
30    if(depth[f1]<depth[f2]){
31        par[f1]=f2;
32    }else{
33        par[f2]=f1;
34        if(depth[f1]==depth[f2]){
35            depth[f1]++;
36            //记录深度
37        }
38    }
39 }
40 
41 bool same(int x,int y){
42     return find_father(x)==find_father(y);
43 }
44 
45 int main()
46 {
47     int t;
48     int n,k;
49     int a,b;
50     bool ans=false;
51     scanf("%d",&t);
52     for(int kk=0;kk<t;kk++){
53         scanf("%d %d",&n,&k);
54         init(n*2);
55         ans=false;
56         for(int i=0;i<k;i++){
57             scanf("%d %d",&a,&b);
58             if(same(a,b)){
59                 ans=true;
60             }else{
61                 unite(a,b+n);
62                 unite(a+n,b);
63             }
64         }
65         printf("Scenario #%d:\n",kk+1);
66         if(ans){
67             printf("Suspicious bugs found!\n");
68         }else{
69             printf("No suspicious bugs found!\n");
70         }
71         printf("\n");
72 
73     }
74     return 0;
75 }

 

转载于:https://www.cnblogs.com/TWS-YIFEI/p/6039296.html