Submission #444213

#	Time	Username	Problem	Language	Result	Execution time	Memory
444213		zaneyu	Collecting Stamps 3 (JOI20_ho_t3)	C++14	0 / 100	1 ms	332 KiB

ho_t3

/*input
6 25
3 4 7 17 21 23
11 7 17 10 8 10
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=double;
using pii=pair<ll,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=1e18+1;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
inline ll mult(ll a,ll b){
    if(a>=MOD) a%=MOD;
    if(b>=MOD) b%=MOD;
    return (a*b)%MOD;
}
inline ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=200+5;
const int maxlg=__lg(maxn)+2;
int dp[maxn][maxn][maxn][2];
int arr[maxn],t[maxn];
int l;
int get(int a,int b){
    return min(abs(arr[a]-arr[b]),l-abs(arr[a]-arr[b]));
}
int32_t main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n;
    cin>>n>>l;
    REP1(i,n) cin>>arr[i];
    REP1(i,n) cin>>t[i];
    REP(i,n+2) REP(j,n+2) REP(k,n+2) REP(z,2) dp[i][j][k][z]=INF;
    dp[0][n+1][0][0]=dp[0][n+1][0][1]=0;
    int ans=0;
    REP(i,n+1){
        for(int j=n+(i==0);j>=i;j--){
            REP(k,n+1){
                //cout<<dp[i][j][k][0]<<' '<<dp[i][j][k][1]<<'\n';
                if(dp[i][j][k][0]!=INF or dp[i][j][k][1]!=INF) MXTO(ans,k);
                if(i==j) continue;
                int t1=min(dp[i][j][k][0]+get(i,i+1),dp[i][j][k][1]+get(i+1,j)),t2=min(dp[i][j][k][1]+get(j,j-1),dp[i][j][k][0]+get(i,j-1));
                MNTO(dp[i+1][j][k+(t1<=t[i+1])][0],t1);
                MNTO(dp[i][j-1][k+(t2<=t[j-1])][1],t2);
            }
        }
    }
    cout<<ans;
}  

• Subtask 1: 0 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 0 / 75