hdu 1829 (并查集)

来源:互联网 发布:大数据具体应用 编辑:程序博客网 时间:2024/06/05 18:28

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
题意:有多个测试组,每个测试组,第一行n,m,是人数n,和接下来m个数对,每给定一数对,每对数表示两个人是相互喜欢的,最后判断有没有同性恋;
思路:首先我们可以分为男女两组,即a表示男,1+maxn表示女,我们用并查集标记关系,最后只需判断同性直接是否有关联就行了;
代码:

#include<cstdio>#include<cstring>#include<algorithm>#define maxn 2000using namespace std;int a[2*maxn+5],n,m;bool flag;int find(int t){    if(t!=a[t])       a[t]=find(a[t]);    return a[t];}void mark(int x,int y){    int zx=find(x);    int zy=find(y-maxn);    if(zx==zy)    {        flag=true;        return;    }    zy=find(y);    if(zx!=zy)      a[zx]=zy;}int main(){    int t,s=1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)           a[i]=i,a[i+maxn]=i+maxn;        flag=false;        for(int i=1;i<=m;i++)        {            int za,zb;            scanf("%d%d",&za,&zb);            if(flag)               continue;            mark(za,zb+maxn);            mark(zb,za+maxn);        }        printf("Scenario #%d:\n",s++);        if(flag)           printf("Suspicious bugs found!\n\n");        else           printf("No suspicious bugs found!\n\n");    }}
0 0
原创粉丝点击