POJ – 2492 A Bug's Life
- 2020 年 4 月 9 日
- 筆記
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!
Problem solving: 這道題的意思就是給你幾組數，每組數有兩個數代表這兩個數，代表這兩個編號的蟲子是相愛的，即他們不是同性的。 問你輸入的數據中會不會存在相同性別（同性戀）的情況存在。
這道題的做法還是很多的，二分圖，帶權並查集什麼什麼的。 但是上面那兩個我都不熟悉。所以選擇了裸並查集加一個小處理的辦法。
這裡我們用x+n代表跟x性別不同的蟲子，每次join的時候對（x+n,y）和(x,y+n)進行join即可。然後如果出現性別相同的直接標記一下，就是並查集啦。因為這裡性別是有兩種，說明是要分成兩隊的，直接並查集顯然是不可以的。
Code:
#include <iostream> using namespace std; #define zhengshu int const int maxn = 1e5; int p[maxn]; int find(int x) { return p[x] != x ? p[x] = find(p[x]) : x; } void join(zhengshu x, zhengshu y) { x = find(x), y = find(y); if (x != y) p[x] = y; } int main() { zhengshu t, n, m, x, y; scanf("%d", &t); for (zhengshu j = 1; j <= t; j++) { zhengshu flag = 0; scanf("%d %d", &n, &m); for (zhengshu i = 1; i <= 2 * n; i++) p[i] = i; for (zhengshu i = 0; i < m; i++) { scanf("%d %d", &x, &y); if (find(x) == find(y)) flag = 1; else join(x + n, y), join(x, y + n); } printf("Scenario #%d:n", j); if (flag) puts("Suspicious bugs found!"); else puts("No suspicious bugs found!"); puts(""); } }
