题解 P4799 【[CEOI2015 Day2]世界冰球锦标赛】

题解 P4799 【[CEOI2015 Day2]世界冰球锦标赛】

双向搜索好题

传送门

实际上，双向搜索就是把\(a^n\)的复杂度转变成了大多为\(O(nlogna^{\frac{n}{2}})\)的复杂度。

上代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<bitset>
#include<vector>
#include<map>
#include<ctime>
#include<cstdlib>
#include<set>
#include<bitset>
#include<stack>
#include<list>
#include<cmath>
using namespace std;
#define RP(t,a,b) for(register int (t)=(a),edd_=(b);t<=edd_;++t)
#define DRP(t,a,b) for(register int (t)=(a),edd_=(b);t>=edd_;--t)
#define ERP(t,a) for(int t=head[a];t;t=e[t].nx)
#define Max(a,b) ((a)<(b)?(b):(a))
#define Min(a,b) ((a)<(b)?(a):(b))
#define TMP template<class ccf>
#define lef L,R,l,mid,pos<<1
#define rgt L,R,mid+1,r,pos<<1|1
#define midd register int mid=(l+r)>>1
#define chek if(R<l||r<L)return
#define all 1,n,1
#define pushup(x) seg[(x)]=seg[(x)<<1]+seg[(x)<<1|1]
typedef long long ll;
TMP inline ccf qr(ccf k){
    char c=getchar();
    ccf x=0;
    int q=1;
    while(c<48||c>57)
	q=c==45?-1:q,c=getchar();
    while(c>=48&&c<=57)
	x=x*10+c-48,c=getchar();
    if(q==-1)
	x=-x;
    return x;
}
const int maxn=2097155;

ll data[45];
ll ord[maxn];
ll ans;
ll n;
ll m;int cnt;
inline ll C(ll k){
    ll mid,l=1,r=cnt;
    do{
	mid=(l+r)>>1;
	if(ord[mid]<=k)
	    l=mid+1;
	else
	    r=mid-1;
    }while(l<=r);
    return l;
}

void dfs(int now,ll rest){
    if(now>(n>>1))return void(ord[++cnt]=m-rest);
    if(rest>=data[now])
	dfs(now+1,rest-data[now]);
    dfs(now+1,rest);
}

void dfs2(int now,ll rest){
    if(now>n) return void(ans+=C(rest)-1ll);
    if(rest>=data[now])
	dfs2(now+1,rest-data[now]);
    dfs2(now+1,rest);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in.in","r",stdin);
    freopen("out.out","w",stdout);
#endif
    n=qr(1ll);
    m=qr(1ll);
    RP(t,1,n)
	data[t]=qr(1ll);
    dfs(1,m);
    sort(ord+1,ord+cnt+1);
    dfs2(n/2+1,m);
    printf("%lld\n",ans);
    return 0;
}