HDUoj 1829 A Bug's Life ( 并查集

A Bug’s Life

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14468 Accepted Submission(s): 4722

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

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!
带权并查集的应用
题意：判断虫子是否是同性恋233333

#include<cstdio>#include<cstring>int par[2020];int relation[2020];bool flag;void init(int n){//  memset(relation,0,sizeof(relation));    for(int i = 1;i <= n; i++) {        par[i] = i;        relation[i] = 0;    }}int find(int x){    if(x == par[x]) {        return x;    }    int t = find(par[x]);    relation[x] = (relation[par[x]] + relation[x]) % 2;  //找到该点父节点与新节点与父节点的关系     par[x] = t;    return par[x];   // 根节点与该节点的关系 }void unite(int x,int y){    int v = find(x);    int u = find(y);    if(u != v) {        relation[v] = (relation[x] + relation[y] +1) % 2;        par[v] = u;    }    else {        if(relation[x] == relation[y])         flag = false;    }} int main(){    int T, k = 1;    scanf("%d",&T);    while(T--) {        flag = true;        int n, m;        scanf("%d%d",&n,&m);        init(n);        int a, b;        for(int i = 0;i < m; i++) {            scanf("%d%d",&a,&b);            unite(a,b);        }        printf("Scenario #%d:\n",k++);        if(flag) {            printf("No suspicious bugs found!\n");        }        else {            printf("Suspicious bugs found!\n");        }        printf("\n");    }return 0;}