[BZOJ 2654] tree

tree

参考博客

https://blog.csdn.net/y752742355/article/details/85468855

https://www.cnblogs.com/cjyyb/p/9438101.html

BZOJ 2654

题目大意

给出一个无向图，每个边有一条边权以及一种颜色(0/1)，问恰好有 \(k\) 条颜色为 \(0\) 的边的最小生成树，保证有解

数据范围

\(V\le50000,E\le100000\)

时空限制

30sec,512MB

分析

设 \(F(x)\) 为包含 \(x\) 条 \(0\) 边的最小生成树大小，显然 \(F(x)\) 的斜率递增，而题目求最小值，符合这两个条件，那么就可以套用带权二分解决此题

注意 \(kruskal\) 过程中要求出答案上界，将边权相同的 \(0\) 边排在前面

Code

#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
inline char nc() {
	static char buf[100000], *l = buf, *r = buf;
	return l==r&&(r=(l=buf)+fread(buf,1,100000,stdin),l==r)?EOF:*l++;
}
template<class T> void read(T & x) {
	x = 0; int f = 1, ch = nc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=nc();}
	while(ch>='0'&&ch<='9'){x=x*10-'0'+ch;ch=nc();}
	x *= f;
}
const int maxn = 50000 + 5;
const int maxm = 100000 + 5;
const int maxv = 100 + 5;
int n, m, k;
int MST, mid;
struct data {
	int u, v, w, c;
	bool operator < (const data & other) const {
		int a = w + (c ? 0 : mid);
		int b = other.w + (other.c ? 0 : mid);
		if(a != b) return a < b;
		return c < other.c;
	}
} e[maxm];
struct union_set {
	int fa[maxn];
	void init() {
		for(int i = 1; i <= n; ++i) fa[i] = i;
	}
	int find(int a) {
		return a == fa[a] ? a : fa[a] = find(fa[a]);
	}
	bool merge(int a, int b) {
		a = find(a), b = find(b);
		if(a == b) return 0;
		fa[a] = b; return 1;
	}
} us;
bool judge() {
	sort(e + 1, e + m + 1);
	us.init();
	int cnt = 0; MST = 0;
	for(int i = 1; i <= m; ++i) {
		int u = e[i].u, v = e[i].v;
		if(us.merge(u, v)) {
			MST += e[i].w + (e[i].c ? 0 : mid);
			if(!e[i].c) cnt++;
		}
	}
	return cnt >= k;
}
int solve() {
	int l = -maxv, r = maxv, re;
	while(l <= r) {
		mid = (l + r) >> 1;
		if(judge()) l = mid + 1, re = MST - k * mid;
		else r = mid - 1;
	}
	return re;
}
int main() {
//	freopen("testdata.in", "r", stdin);
	read(n), read(m), read(k);
	for(int i = 1; i <= m; ++i) {
		read(e[i].u), read(e[i].v), read(e[i].w), read(e[i].c);
		e[i].u++, e[i].v++;
	}
	printf("%d\n", solve());
	return 0;
}