BZOJ 3872 Ant colony

Description

There is an entrance to the ant hill in every chamber with only one corridor leading into (or out of) it. At each entry, there are \(g\) groups of \(m_{1},m_{2},...,m_{g}\) ants respectively. These groups will enter the ant hill one after another, each successive group entering once there are no ants inside. Inside the hill, the ants explore it in the following way:

Upon entering a chamber with \(d\) outgoing corridors yet unexplored by the group, the group divides into \(d\) groups of equal size. Each newly created group follows one of the \(d\) corridors. If \(d=0\), then the group exits the ant hill.

If the ants cannot divide into equal groups, then the stronger ants eat the weaker until a perfect division is possible. Note that such a division is always possible since eventually the number of ants drops down to zero. Nothing can stop the ants from allowing divisibility - in particular, an ant can eat itself, and the last one remaining will do so if the group is smaller than \(d\).

The following figure depicts \(m\) ants upon entering a chamber with three outgoing unexplored corridors, dividing themselves into three (equal) groups of (\(\lfloor \frac{m}{3} \rfloor\)) ants each.

A hungry anteater dug into one of the corridors and can now eat all the ants passing through it. However, just like the ants, the anteater is very picky when it comes to numbers. It will devour a passing group if and only if it consists of exactly \(k\) ants. We want to know how many ants the anteater will eat.

给定一棵有\(n\)个节点的树。在每个叶子节点，有\(g\)群蚂蚁要从外面进来，其中第\(i\)群有\(m[i]\)只蚂蚁。这些蚂蚁会相继进入树中，而且要保证每一时刻每个节点最多只有一群蚂蚁。这些蚂蚁会按以下方式前进：

·在即将离开某个度数为\(d+1\)的点时，该群蚂蚁有\(d\)个方向还没有走过，这群蚂蚁就会分裂成\(d\)群，每群数量都相等。如果\(d=0\)，那么蚂蚁会离开这棵树。

·如果蚂蚁不能等分，那么蚂蚁之间会互相吞噬，直到可以等分为止，即一群蚂蚁有\(m\)只，要分成\(d\)组，每组将会有\(\lfloor \frac{m}{d} \rfloor\)只，如下图。

一只饥饿的食蚁兽埋伏在一条边上，如果有一群蚂蚁通过这条边，并且数量恰为\(k\)只，它就会吞掉这群蚂蚁。请计算一共有多少只蚂蚁会被吞掉。

Input

The first line of the standard input contains three integers \(n, g, k\) (\(2 \le n,g \le 1000000,1 \le k \le 10^9\)), separated by single spaces. These specify the number of chambers, the number of ant groups and the number of ants the anteater devours at once. The chambers are numbered from \(1\) to \(n\).

The second line contains g integers \(m[1],m[2],...,m[g]\)(\(1 \le m[i] \le 10^{9}\)), separated by single spaces, where \(m[i]\) gives the number of ants in the \(i\)-th group at every entrance to the ant hill. The \(n-1\) lines that follow describe the corridors within the ant hill; the \(i\)-th such line contains two integers \(a[i],b[i] (1 \le a[i],b[i] \le n)\), separated by a single space, that indicate that the chambers no.\(a[i]\) and \(b[i]\) are linked by a corridor. The anteater has dug into the corridor that appears first on input.

第一行包含三个整数\(n,g,k\)，表示点数、蚂蚁群数以及\(k\)。

第二行包含\(g\)个整数\(m[1],m[2],...,m[g]\)，表示每群蚂蚁中蚂蚁的数量。

接下来\(n-1\)行每行两个整数，表示一条边，食蚁兽埋伏在输入的第一条边上。

Output

Your program should print to the standard output a single line containing a single integer: the number of ants eaten by the anteater.

一个整数，即食蚁兽能吃掉的蚂蚁的数量。

Sample Input

7 5 3

3 4 1 9 11

1 2

1 4

4 3

4 5

4 6

6 7

Sample Output

21

一道比较水的题目吧（我都想得出，还不水吗？？？）

从食蚁兽潜伏的那两条线段开始bfs，初步推出食蚁兽能吃掉\(k\)只蚂蚁，需要有几只蚂蚁到达该点（该范围一定是一段连续的区间）。递推到叶子之后统计答案即可（lower_bound区间左端点\(l\)，upper_bound区间右端点\(r\)，\(r-l\)即为对于答案的贡献）。

#include<queue>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
using namespace std;

typedef long long ll;
#define maxn (1000010)
int side[maxn],toit[maxn*2],next[maxn*2],m[maxn],g,k,n;
int S,T,cnt,d[maxn];
bool in[maxn]; ll arr[maxn][2],ans,inf;

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

inline void add(int a,int b)
{
	next[++cnt] = side[a]; side[a] = cnt;
	toit[cnt] = b; ++d[a];
}

inline void ins(int a,int b) { add(a,b); add(b,a); }

inline void bfs()
{
	queue <int> team;
	arr[S][0] = (ll)k*(ll)max((d[S]-1),1); arr[S][1] = min((ll)(k+1)*(ll)max((d[S]-1),1)-1LL,inf);
	arr[T][0] = (ll)k*(ll)max((d[T]-1),1); arr[T][1] = min((ll)(k+1)*(ll)max((d[T]-1),1)-1LL,inf);
	in[S] = in[T] = true;
	if (arr[S][0] < inf) team.push(S); if (arr[T][0] < inf) team.push(T);
	while (!team.empty())
	{
		int now = team.front(); team.pop();
		if (d[now] == 1)
		{
			int l = lower_bound(m+1,m+g+1,arr[now][0])-m,r = upper_bound(m+1,m+g+1,arr[now][1])-m;
			ans += (ll)(r-l)*(ll)k;
		}
		for (int i = side[now];i;i = next[i])
		{
			int v = toit[i];
			if (in[toit[i]]) continue;
			in[toit[i]] = true;
			arr[v][0] = arr[now][0]*(ll)(max(d[v]-1,1));
			arr[v][1] = min((arr[now][1]+1LL)*(ll)(max(d[v]-1,1))-1LL,inf);
			if (arr[v][0] < inf) team.push(v);
		}
	}
}

int main()
{
	freopen("3872.in","r",stdin);
	freopen("3872.out","w",stdout);
	n = read(); g = read(); k = read();
	for (int i = 1;i <= g;++i) m[i] = read();
	sort(m+1,m+g+1); inf = m[g]+1;
	S = read(); T = read();
	ins(S,T);
	for (int i = 2;i < n;++i) ins(read(),read());
	bfs(); printf("%lld",ans);
	fclose(stdin); fclose(stdout);
	return 0;
}