CF444E. DZY Loves Planting

题目链接

CF444E. DZY Loves Planting

题解

可以..二分网络流

可是

考虑边从小到大排序

考虑每条边能否成为答案

用并查集维护节点之间的联通性

对于一条边来说，如果这条边可以成为答案

那么当前已经合并的每个点，都需要给它分配一个未被合并的点

这就很好判定了

代码

#include<ctime>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#define rep(a,b,c) for(int a = b;a <= c;++ a)
#define per(a,b,c) for(int a = b;a >= c;-- a)
#define gc getchar()
#define pc putchar
using namespace std;
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c > '9') { if(c == '-')f = - 1 ;c = gc; }
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
	return x * f;
}
#define LL long long
void print(int x) {
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0');
}
const int maxn = 100007;
struct node {
	int u,v,w;
	bool operator < (const node&p)const {
	return w < p.w;
	}	

} e[maxn << 1];
int a[maxn],sum = 0,sz[maxn],fa[maxn];
bool judge = true;
int find(int x) {
	if(fa[x] != x) fa[x] = find(fa[x]);
	return fa[x];
}
void merge(int x,int y) {
	x = find(x),y = find(y);
	sz[x] += sz[y];
	a[x] += a[y];
	fa[y] = x;
	if(sz[x] > sum - a[x]) judge = 0;
}
int main() {
	int n = read();
	for(int i = 1;i < n;++ i) e[i].u = read(),e[i].v = read(),e[i].w=read();
	std::sort(e + 1,e + n);
	for(int i = 1;i <= n;++ i)
		a[i] = read(),sz[i] = 1,sum += a[i],fa[i] = i;
	int ans = 0;
	if(n == 1 && a[1] == 1) while(1){};
	for(int i = 1;i < n;++ i) {
		if(judge) ans = e[i].w;
		merge(e[i].u,e[i].v);
	}
	print(ans);
}