POJ 2492 J-A Bug's Life
阿新 • • 發佈:2018-11-04
http://poj.org/problem?id=2492
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
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!
程式碼:
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; int T, N, M; int f[4010]; void init() { for(int i = 1; i <= 2 * N; i ++) f[i] = i; } int Find(int x) { if(x != f[x]) f[x] = Find(f[x]); return f[x]; } int Merge(int x, int y) { int fx = Find(x); int fy = Find(y); if(fx != fy) { f[fx] = fy; } } bool same(int x, int y) { int fx = Find(x); int fy = Find(y); if(fx == fy) return false; else return true; } int main() { scanf("%d", &T); for(int i = 1; i <= T; i ++) { bool flag = true; scanf("%d%d", &N, &M); init(); for(int j = 1; j <= M; j ++) { int a, b; scanf("%d%d", &a, &b); if(same(a, b) || same(a + N, b + N)) { Merge(a + N, b); Merge(a, b + N); } else flag = false; } if(i != 1) puts(""); printf("Scenario #%d:\n", i); if(!flag) printf("Suspicious bugs found!\n"); else printf("No suspicious bugs found!\n"); } return 0; }