csp-s模擬測試99「淘淘摘蘋果,開心的金明,笨小猴」

淘淘摘蘋果

題解

分塊大法好!

每個塊內維護單調上升序列

而後塊之間暴力lower_bound

複雜度$n*\sqrt n log(\sqrt n)$

代碼

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define A 100005
ll t,n,m,t2;
ll bl[A],h[A],mx[410],sz[410];
ll st[410][410];//可是複雜度不對啊!
int main(){
    freopen("TaoPApp.in","r",stdin);
    freopen("TaoPApp.out","w",stdout);
    scanf("%d%d",&n,&m);
    t=sqrt(n);
    for(ll i=1;i<=n;i++){
        scanf("%d",&h[i]);
        bl[i]=(i-1)/t+1;
        t2=max(t2,bl[i]);
    }
    for(ll i=1;i<=n;i++){
        ll belong=bl[i];
        mx[belong]=max(mx[belong],h[i]);
        if(!sz[belong]){
            sz[belong]++;
            st[belong][sz[belong]]=h[i];
            continue ;
        }
        else if(h[i]>st[belong][sz[belong]]){
            sz[belong]++;
            st[belong][sz[belong]]=h[i];
        }
        
    }
    for(ll i=1;i<=m;i++){
        ll pla,hnow,lst;
        scanf("%d%d",&pla,&hnow);
        ll belong=bl[pla],maxnow=0,cnt=0;
        lst=h[pla];h[pla]=hnow;
        for(ll j=1;j<=t2;j++){
            if(j==belong){
                for(ll k=(j-1)*t+1;k<=min(j*t,n);k++){
                    if(h[k]>maxnow){
                        maxnow=h[k];
                        cnt++;
                    }
                }
            }
            else {
                if(mx[j]>maxnow){
                    cnt+=(sz[j]-(upper_bound(st[j]+1,st[j]+sz[j]+1,maxnow)-st[j])+1);
                    maxnow=mx[j];
                }
            }
        }
        h[pla]=lst;
        printf("%d\n",cnt);
    }
}

View Code

或者能夠線段樹

代碼

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define A 5555555
struct node{
    ll l,r,max,val;
}tr[A];
ll a[A];
ll n,m,lst;
void built(ll x,ll l,ll r){
    tr[x].l=l,tr[x].r=r;
    if(l==r){
        return ;
    }
    ll mid=(l+r)>>1;
    built(x<<1,l,mid);
    built(x<<1|1,mid+1,r);
}
ll ask(ll x,ll val){
    if(tr[x].l==tr[x].r)    return tr[x].max>val;
    ll mid=(tr[x].l+tr[x].r)>>1;
    if(tr[x<<1].max<=val) return ask(x<<1|1,val);
    else return ask(x<<1,val)+tr[x].val-tr[x<<1].val;
}
void up(ll x){
    tr[x].max=max(tr[x<<1].max,tr[x<<1|1].max);
    tr[x].val=tr[x<<1].val+ask(x<<1|1,tr[x<<1].max);
}
void change(ll x,ll pla,ll val){
    if(tr[x].l==tr[x].r){
        tr[x].val=1;
        tr[x].max=a[pla];
        return ;
    }
    ll mid=(tr[x].l+tr[x].r)>>1;
    if(mid>=pla) change(x<<1,pla,val);
    else change(x<<1|1,pla,val);
    up(x);
}
int main(){

    freopen("TaoPApp.in","r",stdin);
    freopen("TaoPApp.out","w",stdout);
    scanf("%lld%lld",&n,&m);
    built(1,1,n);
    for(ll i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        change(1,i,a[i]);
    }
    for(ll i=1;i<=m;i++){
        ll p,h;
        scanf("%lld%lld",&p,&h);
        lst=a[p];
        a[p]=h;
        change(1,p,a[p]);
        printf("%lld\n",tr[1].val);
        a[p]=lst;
        change(1,p,a[p]);
    }
}

View Code

然而分塊比線段樹快的多

開心的金明

題解

dp不可作就要往貪心\二分上想了

貪心考慮每原材料

    val[1]=s1[1].c;
    for(ll i=2;i<=k;i++){
        val[i]=min(val[i-1]+s2[i-1].R,s1[i].c);
    }

咱們先把全部電腦生產出來,等賣出時再計算生產消耗,能生產就生產

貪心賣就是賣出生產代價最小的

證實:若是最後生產代價大生產代價小都賣出去了,那麼和先賣生產代價小再賣生產代價大,狀況同樣

若是生產代價小沒賣出去,而賣出生產代價大的確定不優

因而拿個堆維護一下

這種費用滯後計算思想仍是要見一見

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define A 111111
ll read(){
    ll x;scanf("%lld",&x);return x;
}
ll tot,n,k,tot2,ans;
ll val[A];
struct node{
    ll val,sum;
    node(){}
    node(const ll &a,const ll &b){val=a;sum=b;}
    friend bool operator < (const node &a,const node &b){
        return a.val<b.val;
    }
}tmp[A],now[A];
struct some{
    ll c;//原材料價格
    ll d;//至少d個電腦
    ll m;//生產電腦費用
    ll p;//能夠生產p臺
    ll e;//最多儲存e臺
    ll R;//儲存消耗R
    ll E;//儲存電腦消耗
}s1[A],s2[A];
int main(){
    freopen("happy.in","r",stdin);
    freopen("happy.out","w",stdout);
    scanf("%lld",&k);
    for(ll i=1;i<=k;i++){
        scanf("%lld%lld%lld%lld",&s1[i].c,&s1[i].d,&s1[i].m,&s1[i].p);
    }
    for(ll i=1;i<=k-1;i++){
        scanf("%lld%lld%lld",&s2[i].e,&s2[i].R,&s2[i].E);
    }
    val[1]=s1[1].c;
    for(ll i=2;i<=k;i++){
        val[i]=min(val[i-1]+s2[i-1].R,s1[i].c);
    }
    for(ll i=1;i<=k;i++){
        for(ll j=1;j<=tot;j++)
            tmp[j].val+=s2[i-1].E;
        tmp[++tot]=node(val[i]+s1[i].m,s1[i].p);
//        printf("s1.d=%lld\n",s1[i].d);
        sort(tmp+1,tmp+tot+1);
        ll ned=s1[i].d,cun=s2[i].e;
        for(ll j=1;j<=tot;j++){
            if(tmp[j].sum<=ned){
                ans+=tmp[j].sum*tmp[j].val;
//                printf("i=%lld val=%lld sum=%lld\n",i,tmp[j].val,tmp[j].sum);
                ned-=tmp[j].sum;
                tmp[j].sum=0;
                
            }
            else {
                ans+=ned*tmp[j].val;
//                printf("val=%lld ned=%lld\n",tmp[j].val,ned);
                tmp[j].sum-=ned;
                ned=0;
            }
        }
        if(ned){
            puts("-1");
            return 0;
        }
//        printf("ans=%lld\n",ans);
        tot2=0;
        for(ll j=1;j<=tot;j++){
            if(tmp[j].sum==0) continue ;
            if(tmp[j].sum<=cun){
                now[++tot2]=tmp[j];
                cun-=tmp[j].sum;
            }
            else {
                now[++tot2]=node(tmp[j].val,cun);
                cun=0;
            }
        }
        tot=tot2;
        for(ll j=1;j<=tot;j++)
            tmp[j]=now[j];
    }
    printf("%lld\n",ans);
}

View Code

笨小猴

題解

發現check是O1的(維護前綴和)然而可能初始給定序列根本組不成

而後能夠每random_shuffle1次再check500000次

進行上述操做50次就AC了

代碼

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define A  1111111 
struct node{
    ll a,b,id;
}c[A];
ll suma[A],sumb[A],sa,sb,n;
ll random(ll x){
    return rand()%x;
}
int main(){
    freopen("grandmaster.in","r",stdin);
    freopen("grandmaster.out","w",stdout);
    srand((unsigned)time(0));
    scanf("%lld",&n);
    for(ll i=1;i<=2*n+1;i++){
        scanf("%lld%lld",&c[i].a,&c[i].b);
        c[i].id=i;
        sa+=c[i].a,sb+=c[i].b;
    }
    for(ll i=1;i*n<=1000000;i++){
        random_shuffle(c+1,c+2*n+2);
        for(ll j=1;j<=2*n+1;j++)
            suma[j]=suma[j-1]+c[j].a,
            sumb[j]=sumb[j-1]+c[j].b;
        for(ll j=1;j<=n;j++){
            ll l=random(n+1)+1;
            if(suma[l+n]-suma[l-1]>sa-(suma[l+n]-suma[l-1]))
                if(sumb[l+n]-sumb[l-1]>sb-(sumb[l+n]-sumb[l-1])){
                    for(ll k=l;k<=l+n;k++)
                        printf("%lld\n",c[k].id);
                    return 0;
                }
        }
        if(clock()>990000){
            printf("-1\n");
            return 0;
        }
    }
}

View Code