A Bug's Life
| Time Limit: 10000MS | Memory Limit: 65536K | |
| Total Submissions: 35520 | Accepted: 11650 |
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!
Hint
Huge input,scanf is recommended.趁熱打鐵，馬上找了帶權并查集的題來做一做。
這個題就是告訴你兩個蟲子的性別不一樣，然后問你存不存在關系產生矛盾的蟲子。像第一組示例，1、2不同，2、3不同，那么1 3 應該是同性，結果又出來個1 3.所以產生了矛盾。
比較基礎的帶權并查集，也是兩個狀態。
沒學過帶權并查集的可以先看下這里：點擊打開鏈接
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN=2010;int n,m;int pre[MAXN],relation[MAXN];int findx(int x){ if(pre[x]==x)return x; int order=pre[x]; pre[x]=findx(pre[x]); relation[x]=(relation[x]+relation[order])%2; return pre[x];}int main(){ int i; int t; scanf("%d",&t); for(int tt=1; tt<=t; ++tt) { scanf("%d%d",&n,&m); int x,y; int flag=0; for(i=1;i<=n;++i) { pre[i]=i; relation[i]=0; } for(i=1; i<=m; ++i) { scanf("%d %d",&x,&y); if(flag)continue; int a=findx(x),b=findx(y); if(a!=b) { pre[b]=a; relation[b]=(relation[x]-relation[y]+1)%2; } else { int p=(relation[x]-relation[y]+2)%2; if(!p) { flag=1; } } } printf("Scenario #%d:/n",tt); if(flag)puts("Suspicious bugs found!/n"); else puts("No suspicious bugs found!/n"); } return 0;}
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