【BZOJ2654】tree【二分】【最小生成樹】

【題目鏈接】

奇怪的二分。

考慮給白邊的邊權加上一個數，這個數越大，MST時選的白邊就越少。

注意排序時候，如果邊權相等，要先選白邊。

/* Forgive me Not */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 50005, maxm = 100005, inf = 0x3f3f3f3f;

int n, m, k, fa[maxn];

struct edge {
	int u, v, w, c;

	bool operator < (const edge &x) const {
		return w != x.w ? w < x.w : c < x.c;
	}
} g[maxm], e[maxm];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline int find(int x) {
	return fa[x] == x ? x : fa[x] = find(fa[x]);
}

inline int check(int mid) {
	int cnt = 0, res = 0;
	for(int i = 1; i <= n; i++) fa[i] = i;
	for(int i = 1; i <= m; i++) {
		e[i] = g[i];
		if(!e[i].c) e[i].w += mid;
	}
	sort(e + 1, e + 1 + m);
	for(int i = 1; i <= m; i++) {
		int u = find(e[i].u), v = find(e[i].v);
		if(u != v) {
			res += e[i].w;
			if(!e[i].c) cnt++;
			fa[u] = v;
		}
	}
	return cnt >= k ? res : -1;
}

int main() {
	n = iread(); m = iread(); k = iread();
	for(int i = 1; i <= m; i++) {
		int u = iread(), v = iread(), w = iread(), c = iread(); u++; v++;
		g[i] = (edge){u, v, w, c};
	}

	int l = -101, r = 101, ans = inf;
	while(l <= r) {
		int mid = l + r >> 1;
		int res = check(mid);
		if(res != -1) ans = res - k * mid, l = mid + 1;
		else r = mid - 1;
	}

	printf("%d\n", ans);
	return 0;
}