POJ 2492 A Bug's Life
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.
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.
大意就是,题目给出一对又一对的情侣,看看其中是不是有同性恋。
种类并查集。
用一个数组re保存祖先,然后判断两者的祖先是否相等,相等则同性。
//Asimple //#include <bits/stdc++.h> #include <iostream> #include <sstream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <cctype> #include <cstdlib> #include <stack> #include <cmath> #include <set> #include <map> #include <string> #include <queue> #include <limits.h> #include <time.h> #define INF 0xfffffff #define mod 10007 #define CLS(a, v) memset(a, v, sizeof(a)) #define debug(a) cout << #a << " = " << a <<endl #define abs(x) x<0?-x:x using namespace std; typedef long long ll; const int maxn = 2005; int fa[maxn]; int re[maxn]; int n, m, num, T, x, y; int find(int x) { if( fa[x]==x ) return x; int t = fa[x]; fa[x] = find(fa[x]); re[x] = ( re[x] + re[t] +1 )%2; return fa[x]; } void make_set(int x, int y) { int xx = find(x); int yy = find(y); fa[xx] = yy; re[xx] = (re[y]-re[x])%2; } void input() { cin >> T; int cas = 1; while( cas<= T ) { bool f = true; cin >> n >> m; for(int i=1; i<=n; i++){ fa[i] = i; re[i] = 1; } for(int i=0; i<m; i++) { cin >> x >> y; if( find(x)==find(y) ) { if( re[x]==re[y] ) f = false; } else make_set(x, y); } printf("Scenario #%d:\n", cas++); puts(f?"No suspicious bugs found!":"Suspicious bugs found!"); puts(""); } } int main() { input(); return 0; }
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