LightOJ 1009 Back to Underworld (种类并查集)

Posted 2022-11-29 synapse7

http://lightoj.com/volume_showproblem.php?problem=1009

种类并查集实现（当然用二分图染色也可以）

/*0.204s,2820KB*/

#include<bits/stdc++.h>
using namespace std;
const int mx = 20000;

int fa[mx * 2 + 5], rk[mx * 2 + 5], x[100005], y[100005];
bool has[mx * 2 + 5];

int find(int x) return ~fa[x] ? fa[x] = find(fa[x]) : x;

void merge(int x, int y)

	x = find(x), y = find(y);
	if (x == y) return;
	fa[y] = x;
	rk[x] += rk[y];


int main()

	int t, n, ans, cnt, u, v, fau, fav, i;
	scanf("%d", &t);
	for (int cas = 1; cas <= t; ++cas)
	
		memset(fa, -1, sizeof(fa));
		memset(rk, 0, sizeof(rk));
		memset(has, 0, sizeof(has));
		scanf("%d", &n);
		ans = cnt = 0;
		while (n--)
		
			scanf("%d%d", &u, &v);
			x[cnt] = u, y[cnt++] = v;
			has[u] = has[v] = has[u + mx] = has[v + mx] = true;
			rk[u] = rk[v] = 1;
		
		while (cnt--)
		
			merge(x[cnt], y[cnt] + mx);
			merge(x[cnt] + mx, y[cnt]); ///切莫弄反
		
		for (i = 1; i <= mx * 2; ++i)
			if (has[i] && fa[i] == -1)
			
				fau = i, fav = find(i + mx);
				ans += max(rk[fau], rk[fav]);
				fa[fau] = fa[fav] = 0; ///已访问
			
		printf("Case %d: %d\\n", cas, ans);
	
	return 0;