A Bug's Life （分组 并查集）

A Bug's Life

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1087 Accepted Submission(s): 399

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.
 

 题意：   每次输入 a ，b   然后 表示a，b表示不同的性别，然后判断每组情况中是否有gay的出现。

很明显的并查集，但是如何判断两者是不是gay呢。。。。可以分两组并查集，a 并 b。然后b 再并 a，如果存在。pre[a] == pre[b]的情况 那就是有 gay 存在了。

#include <stdio.h>#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string.h>using namespace std;const int MAX = 2000;int pre[(MAX<<1) + 5];bool flag;void init() {    for (int i = 1; i<= (MAX<<1) + 5; ++i)        pre[i] = i;    flag = 0;}int find(int x) {    if (pre[x] == x)        return x;    pre[x] = find(pre[x]);    return pre[x];}void merge(int x, int y) {    int prex = find(x);    int prey = find(y - MAX);    if (prex == prey) {        flag = 1;        return ;    }    prey = find(y);        if (prex != prey)        pre[prey] = pre[prex];}int main() {    int Case = 1;    int T;    int n, m;    cin >> T;    int a, b;    while (T--) {        memset(pre, 0, sizeof(pre));        init();        scanf("%d %d", &n, &m);        for (int i = 0; i < m; ++i) {            scanf("%d %d", &a, &b);            if (flag)                continue;            merge(a, b + MAX);            merge(b, a + MAX);        }        printf("Scenario #%d:\n", Case++);        if (flag)            printf("Suspicious bugs found!\n");        else            printf("No suspicious bugs found!\n");               printf("\n");    }}