思路：保存每個(gè)點(diǎn)與其父節(jié)點(diǎn)的關(guān)系，注意合并和路徑壓縮即可。

AC代碼

#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 2000 + 5;
struct node{
	int par;
	int real; //0表示同性,1表示性別相異
}a[maxn];

int find(int x, int &root) {
	if(a[x].par == x) {
		root = x;
		return a[x].real;
	}
	int r = (a[x].real + find(a[x].par, root)) % 2;
	a[x].par = root;
	return a[x].real = r;
}

bool unionset(int x, int y) {
	int r1, r2;
	int rx = find(x, r1), ry = find(y, r2);
	if(r1 == r2) {
		if(rx == ry) return false;
	}
	else {
		a[r1].par = y;
		a[r1].real = (rx + 1) % 2;
	}
	return true;
}

int main() {
	int n, m, T, kase = 1;
	scanf("%d", &T);
	while(T--) {
		if(kase > 1) printf("\n");
		scanf("%d%d", &n, &m);
		for(int i = 0; i <= n; ++i) {
			a[i].par = i;
			a[i].real = 0;
		}
		int x, y;
		int flag = 1;
		for(int i = 0; i < m; ++i) {
			scanf("%d%d", &x, &y);
			if(flag && !unionset(x, y)) {
				flag = 0;
			}
		}
		printf("Scenario #%d:\n", kase++);
		if(flag) printf("No suspicious bugs found!\n");
		else printf("Suspicious bugs found!\n");
	}
	return 0;
}

如有不當(dāng)之處歡迎指出！

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