HDOJ1829 A Bug's Life(并查集)

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!
Hint
Huge input,scanf is recommended.
 

 

和食物链类似，均是种类并查集，找relation很重要，1为同性，0为异性。

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 2005;int par[maxn], relation[maxn];int find(int x){if(par[x] == x) return x;int y = par[x];par[x] = find(par[x]);relation[x] = (relation[x] + relation[y] + 1) % 2;return par[x];}void merge(int a, int b){int x = find(a), y = find(b);par[x] = y;relation[x] = (relation[b] - relation[a]) % 2;}int main(int argc, char const *argv[]){int t, cas = 1;scanf("%d", &t);while(t--) {bool flag = false;int n, m;scanf("%d%d", &n, &m);for(int i = 1; i <= n; ++i) {par[i] = i;relation[i] = 1;}for(int i = 0; i < m; ++i) {int a, b;scanf("%d%d", &a, &b);if(find(a) == find(b)) {if(relation[a] == relation[b]) flag = true;}else merge(a, b);}if(flag) printf("Scenario #%d:\nSuspicious bugs found!\n\n", cas++);else printf("Scenario #%d:\nNo suspicious bugs found!\n\n", cas++);}return 0;}