1489 蜥蜴和地下室

思路：先把最左和最右的干掉，剩下的再一起dfs。dfs的时候注意当第cur-1个死了之后（而第cur个不要求必须死）才向后面搜索第cur+1个。

当前的第cur个可以被扔在自己身上的火球，也可被扔在第cur+1个的火球干掉。

#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <sstream>
#include <time.h>
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
#define mt(A,B) memset(A,B,sizeof(A))
#define mod 1000000007
using namespace std;
typedef long long LL;
const int N=2e5+10;
const int inf = 0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3fLL;
int x[N];
int n,a,b,ans,res_2=inf;


void dfs(int cur,int num)
{
    int p=0,q;
    if(cur==n-1)
    {
        ans=min(ans,num);
        return;
    }
    if(x[cur-1]<0)dfs(cur+1,num);

    if(x[cur-1]>=0)
    {
        p=x[cur-1]/b+1;
        x[cur-1]-=p*b;
        x[cur]-=p*a;
        x[cur+1]-=p*b;
        dfs(cur+1,num+p);
        x[cur-1]+=p*b;
        x[cur]+=p*a;
        x[cur+1]+=p*b;
    }
    q=x[cur]/a+1;
    if(x[cur]>=0&&q>p)
    {
        for(int i=p+1;i<=q;i++)
        {
            x[cur-1]-=i*b;
            x[cur]-=i*a;
            x[cur+1]-=i*b;
            dfs(cur+1,num+i);
            x[cur]+=i*a;
            x[cur+1]+=i*b;
            x[cur-1]+=i*b;
        }
    }
}
int main()
{
#ifdef Local
    freopen("data.txt","r",stdin);
#endif
    int p=0,q=0;
    cin>>n>>a>>b;
    ans=inf;
    for(int i=0; i<n; i++)scanf("%d",&x[i]);
    p=x[0]/b+1;
    x[0]-=p*b;;
    x[1]-=p*a;
    x[2]-=p*b;
    if(x[n-1]>=0)
    {
        q=x[n-1]/b+1;
        x[n-2]-=q*a;
        x[n-3]-=q*b;
        x[n-1]-=q*b;
    }
    dfs(1,0);
    if(ans==inf)ans=0;
    cout<<ans+p+q<<endl;
#ifdef Local
    cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
#endif
}