BZOJ2525 [Poi2011]Dynamite 【二分 + 贪心】

题目链接

BZOJ2525

题解

就是要求所有有炸弹的点到点燃点距离最大值最小

显然二分答案距离\(D\)

然后按深度排序，贪心点燃当前没覆盖的深度最深的点往上第\(D\)层的点

每覆盖一个点要标记其能到达的点

显然暴力标记均摊是\(O(n)\)的

复杂度\(O(nlogn)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 300005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
	ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
int n,m,tag[maxn],dep[maxn],fa[maxn],L,tot;
int id[maxn],vis[maxn],ok[maxn],N;
void dfs(int u){
	Redge(u) if ((to = ed[k].to) != fa[u]){
		fa[to] = u; dep[to] = dep[u] + 1;
		dfs(to);
	}
}
void Dfs(int u,int len,int pre){
	ok[u] = true;
	if (vis[u] >= len) return;
	vis[u] = len;
	if (!len) return;
	Redge(u) if ((to = ed[k].to) != pre)
		Dfs(to,len - 1,u);
}
bool check(int mid){
	tot = 0; L = mid;
	REP(i,n) vis[i] = ok[i] = 0;
	REP(i,N){
		int u = id[i],k;
		if (!ok[u]){
			k = L;
			while (k-- && fa[u]) u = fa[u];
			Dfs(u,L,0);
			tot++;
		}
	}
	return tot <= m;
}
inline bool cmp(const int& a,const int& b){
	return dep[a] > dep[b];
}
int main(){
	n = read(); m = read();
	for (int i = 1; i <= n; i++){
		tag[i] = read();
		if (tag[i]) id[++N] = i;
	}
	for (int i = 1; i < n; i++) build(read(),read());
	dfs(1);
	sort(id + 1,id + 1 + N,cmp);
	int l = 0,r = n,mid;
	while (l < r){
		mid = l + r >> 1;
		if (check(mid)) r = mid;
		else l = mid + 1;
	}
	printf("%d\n",l);
	return 0;
}