$bzoj3872\ [Poi2014]\ Ant\ colony$ 二分+$dp$

正解:二分+$dp$

解题报告:

传送门$QwQ$

一年过去了依然没有头绪,,,$gql$的$NOIp$必将惨败了$kk$.

考虑倒推,因为知道知道除数和答案,所以可以推出被除数的范围,然后一路推到叶子节点就成$QwQ$

$over$

嗷注意一个细节是有可能乘爆,所以每次和$m_max$取个$min$就成$QwQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define int long long
#define t(i) edge[i].to
#define ri register int
#define rc register char
#define rb register bool
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];i;i=edge[i].nxt)

const int N=+;
int n,head[N],ed_cnt,g,K,m[N],frx,fry,in[N],l[N],r[N],inf,as;
struct ed{int to,nxt;}edge[N<<];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void ad(ri x,ri y){edge[++ed_cnt]=(ed){x,head[y]};head[y]=ed_cnt;++in[x];}
void dfs(ri nw,ri fa)
{
    e(i,nw)
        if(t(i)^fa)
            l[t(i)]=min(1ll*l[nw]*(in[nw]-),inf+),r[t(i)]=min(1ll*(r[nw]+)*(in[nw]-)-,inf),dfs(t(i),nw);
}
il int fd1(ri x){return lower_bound(m+,m++g,x)-m-;}
il int fd2(ri x){ri tmp=upper_bound(m+,m++g,x)-m;return m[tmp]==x?tmp:tmp-;}

signed main()
{
    freopen("3872.in","r",stdin);freopen("3872.out","w",stdout);
    n=read();g=read();K=read();rp(i,,g)m[i]=read();sort(m+,m++g);inf=m[g];
    rp(i,,n-){ri x=read(),y=read();ad(x,y);ad(y,x);if(i==)frx=x,fry=y;}
    l[frx]=r[frx]=l[fry]=r[fry]=K;dfs(frx,fry);dfs(fry,frx);
    rp(i,,n)if(in[i]==)as+=fd2(r[i])-fd1(l[i]);;printf("%lld\n",1ll*as*K);
    return ;
}