POJ 2492 A Bug's Life——并查集

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题意：两个集合，T组数据，每组数据以N M开头，表示有编号从1到n的n个物品以及m个条件，每个条件给出两个数a b，表示a b不在同一集合，如果所有的a b都不在同一集合那么没有BUG，否则有BUG

思路：参考poj1703，本题是poj1703的简化版

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn =  2 * 1e5 + 10;int T, N, M, par[maxn], ran[maxn];void Init() {    for (int i = 0; i <= (N<<1); i++) par[i] = i, ran[i] = 0;}int Query(int x) { return par[x] == x ? x : par[x] = Query(par[x]); }void Union(int x, int y) {    x = Query(x), y = Query(y);    if (x == y) return;    if (ran[x] < ran[y]) par[x] = y;    else {        par[y] = x;        if (ran[x] == ran[y]) ran[x]++;    }}bool Same(int x, int y) {    return Query(x) == Query(y);}int main() {    scanf("%d", &T);    for (int kase = 1; kase <= T; kase++) {        scanf("%d %d", &N, &M);        Init();        bool ok = true;        while (M--) {            int a, b; scanf("%d%d", &a, &b);            bool ok1 = Same(a, b), ok2 = Same(a + N, b), ok3 = Same(a, b + N);            if (ok1 && !ok2 && !ok3) ok = false;            if (ok) { Union(a + N, b); Union(a, b + N); }        }        printf("Scenario #%d:\n", kase);        if (!ok) printf("Suspicious bugs found!\n\n");        else printf("No suspicious bugs found!\n\n");    }}

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