hdu1829(A Bug's Life)

题目链接:传送门

题目大意:有n个昆虫，有m组关系，接下来m行表示两个昆虫性别不同，问是否有矛盾情况（同男同女）

题目思路:并查集的高级应用，开两倍数组大小，后n个数组表示和当前昆虫不同性别的集合

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <cctype>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <climits>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define fi first
#define se second
#define seg int root,int l,int r
#define ping(x,y) ((x-y)*(x-y))
#define mst(x,y) memset(x,y,sizeof(x))
#define mcp(x,y) memcpy(x,y,sizeof(y))
#define Min(x,y) (x<y?x:y)
#define Max(x,y) (x>y?x:y)
using namespace std;
#define gamma 0.5772156649015328606065120
#define MOD 100000007
#define inf 0x3f3f3f3f
#define N 100005
#define maxn 10001000
typedef long long LL;
typedef pair<int,int> PII;

int fp[N];
int findp(int x){return fp[x]==x?x:fp[x]=findp(fp[x]);}

int main(){
    int i,j,group,k,v,x,y,Case=0,n,m;
    scanf("%d",&group);
    while(group--){
        int flag=0;
        scanf("%d%d",&n,&m);
        for(i=0;i<=n<<1;++i) fp[i]=i;
        for(i=0;i<m;++i){
            scanf("%d%d",&x,&y);
            if(!flag){
                int xa=findp(x);
                int ya=findp(y);
                if(xa==ya) flag=1;
                else{
                    fp[xa]=findp(y+n);
                    fp[ya]=findp(x+n);
                }
            }
        }
        printf("Scenario #%d:\n",++Case);
        if(flag) printf("Suspicious bugs found!\n\n");
        else printf("No suspicious bugs found!\n\n");
    }
    return 0;
}

hdu1829(A Bug's Life)