csp-s模拟测试44「D·E·F」

用心出题,用脚造数据

乱搞场

 1 #include<bits/stdc++.h>
 2 #define re register
 3 #define int long long
 4 #define inf 0x7ffffffffffffff
 5 using namespace std;
 6 int n,a[100010],b[100010],ans=inf;
 7 double st,ed;
 8 inline int read(){
 9     re int a=0,b=1; re char ch=getchar();
10     while(ch<'0'||ch>'9')
11         b=(ch=='-')?-1:1,ch=getchar();
12     while(ch>='0'&&ch<='9')
13         a=(a<<3)+(a<<1)+(ch^48),ch=getchar();
14     return a*b;
15 }
16 inline int max(re int x,re int y){if(x>y) return x; return y;}
17 inline int min(re int x,re int y){if(x<y) return x; return y;}
18 inline void dfs(re int x,re int l,re int r,re int ll,re int rr){
19     if(1ll*(r-l)*(rr-ll)>ans) return ;
20     if(x>n){
21         ans=1ll*(r-l)*(rr-ll); 
22         ed=clock();
23         if((ed-st)/1e6>=1.99){
24             printf("%lld\n",ans);
25             exit(0);
26         }
27         return ; 
28     }
29     dfs(x+1,max(l,a[x]),min(r,a[x]),max(ll,b[x]),min(rr,b[x]));
30     dfs(x+1,max(l,b[x]),min(r,b[x]),max(ll,a[x]),min(rr,a[x]));
31 }
32 signed main(){
33 //    freopen("in.txt","r",stdin);
34     n=read();if(n==1){puts("1");return 0;}
35     for(re int i=1;i<=n;++i){
36         a[i]=read(),b[i]=read();
37         if(a[i]<b[i]) a[i]^=b[i]^=a[i]^=b[i];
38     }
39     st=clock();
40     dfs(1,0,inf,0,inf);
41     printf("%lld\n",ans);
42     return 0;
43 }
随机化
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<bits/stdc++.h>
 6 #define reg register
 7 using namespace std;
 8 typedef long long ll;
 9 const int maxn=1e5+5,INF=2e9;
10 inline void read(int &x)
11 {
12     x=0;char c=getchar();
13     while(c<'0'||c>'9') c=getchar();
14     while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+c-48,c=getchar();
15 }
16 ll ans;
17 int S,T,n,tot,L[maxn][2],R[maxn][2],L2[2],R2[2],tmp[2],yet[maxn];
18 struct ball{
19     int x[2];
20 }c[maxn];
21 struct wh{
22     int x,id,f;
23     bool friend operator < (const wh a,const wh b) {return a.x<b.x;}
24 }g[maxn<<1];
25 void dfs(int x)
26 {
27     T=clock();
28     if(T-S>1500000) {printf("%lld\n",ans);exit(0);}
29     if(x!=1&&1LL*(L[x-1][1]-L[x-1][0])*(R[x-1][1]-R[x-1][0])>ans) return ;
30     if(x==n+1) {ans=1LL*(L[x-1][1]-L[x-1][0])*(R[x-1][1]-R[x-1][0]);return ;}
31     if(c[x].x[0]>c[x].x[1]) swap(c[x].x[0],c[x].x[1]);
32     for(reg int j=0;j<=1;++j)    
33     {
34         L[x][0]=min(L[x-1][0],c[x].x[j]);
35         L[x][1]=max(L[x-1][1],c[x].x[j]);
36         R[x][0]=min(R[x-1][0],c[x].x[j^1]);
37         R[x][1]=max(R[x-1][1],c[x].x[j^1]);
38         dfs(x+1);
39     }
40 }
41 int main()
42 {
43     srand(time(NULL));
44     S=clock();
45 //    freopen("ans.in","r",stdin);
46 //    freopen("b.out","w",stdout);
47     read(n);
48     for(reg int i=1;i<=n;++i)
49     {
50         read(c[i].x[0]),read(c[i].x[1]);
51         g[++tot]=(wh){c[i].x[0],i,0};
52         g[++tot]=(wh){c[i].x[1],i,1};
53     }
54     reverse(c+1,c+n+1);
55     sort(g+1,g+tot+1);
56     ans=1e18;ans+=5;
57     int l=1,r=tot,k=0;
58     memset(yet,0xFF,sizeof(yet));
59     while(k<n)
60     {
61         while(yet[g[l].id]!=-1) ++l;
62         yet[g[l].id]=g[l].f;++k;
63         if(k==n) break;
64         while(yet[g[r].id]!=-1) --r;
65         yet[g[r].id]=g[r].f;++k;
66     }
67     L[0][0]=R[0][0]=L2[0]=R2[0]=INF;
68     for(reg int i=1;i<=n;++i)
69     {
70         L2[0]=min(L2[0],c[i].x[yet[i]]);
71         L2[1]=max(L2[1],c[i].x[yet[i]]);
72         R2[0]=min(R2[0],c[i].x[yet[i]^1]);
73         R2[1]=max(R2[1],c[i].x[yet[i]^1]);
74     }
75     ans=1LL*(L2[1]-L2[0])*(R2[1]-R2[0]);
76     dfs(1);
77     printf("%lld\n",ans);
78     return 0;
79 }
clock

垃圾zzn当然什么也不会啦,乱搞什么也没打

D

考试时想到正解,没打,觉得这仅仅是个简单的剪枝,没想到啊

题意

求$(r-l+1)*gcd(a[l],a[l+1],.....,a[r])$最大值

题解

垃圾zzn没打正解,类正解多$log$用来二分了,常数较小

$gcd$总需要求,求次数太多了

考虑二分,维护$gcd$从$mid$前缀和后缀和

这样你就有$50$分了

考虑$gcd$变化次数小于是维护单调队列,很简单

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define A 333333
 5 ll a[A],suml[A],sumr[A],dll[A],dlr[A];
 6 ll ans=0,n;
 7 ll gcd(ll x,ll y){
 8     if(y==0) return x;
 9     return gcd(y,x%y);
10 }
11 void solve(ll l,ll r){
12     if(l==r) return ;
13     ll mid=(l+r)>>1,rnow=mid+1,lnow=mid;
14     suml[lnow]=a[lnow],sumr[rnow]=a[rnow];
15     dll[0]=0,dlr[0]=0;
16     while(lnow>l){
17         lnow--;
18         suml[lnow]=gcd(suml[lnow+1],a[lnow]);
19         if(suml[lnow]!=suml[lnow+1])
20             dll[++dll[0]]=lnow+1;
21     }
22     dll[++dll[0]]=l;
23     while(rnow<r){
24         rnow++;
25         sumr[rnow]=gcd(sumr[rnow-1],a[rnow]);
26         if(sumr[rnow]!=sumr[rnow-1])
27             dlr[++dlr[0]]=rnow-1;
28     }
29     dlr[++dlr[0]]=r;
30     for(ll lh=1;lh<=dll[0];lh++)
31         for(ll rh=1;rh<=dlr[0];rh++){
32             ll nowl=dll[lh],nowr=dlr[rh];
33             ll g=gcd(suml[nowl],sumr[nowr]);
34 //            printf("nowl=%lld nowr=%lld =%lld\n",dll[lh],dlr[rh],g*(nowr-nowl+1));
35             if(g==1) break;
36             ans=max(ans,g*(nowr-nowl+1));
37         }
38     ans=max(ans,gcd(suml[l],sumr[r])*(r-l+1));
39     solve(l,mid);solve(mid+1,r);
40 }
41 //10 10 101 10 10 
42 int main(){
43 //    freopen("da.in","r",stdin);
44 //    freopen("ans.sol","w",stdout);
45     scanf("%lld",&n);
46     for(ll i=1;i<=n;i++){
47         scanf("%lld",&a[i]);
48     }
49     solve(1,n);
50     printf("%lld\n",ans);
51 }
View Code

 

E

这个题真的很迷

题解

 垃圾zzn考试时打的第二个贪心,然后只有$60$分,事实上单纯第一个贪心就可以$100$分,数据特别水,第一个贪心明明连样例都过不去

没遇到数据这么水的,

正确性垃圾zzn当然不会验证啦

代码也懒的放了

F

题解

0分算法

直接暴力$dp$

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 ll f[2][2019][2019],w[2019];
 5 ll a,b,n,q;
 6 int main(){
 7     scanf("%lld%lld%lld%lld",&n,&q,&a,&b);
 8     memset(f,0x7f,sizeof(f));
 9     for(ll i=1;i<=q;i++)
10         scanf("%lld",&w[i]);
11     f[1][a][w[1]]=abs(b-w[1]);
12     f[1][w[1]][b]=abs(a-w[1]);
13     for(ll i=2;i<=q;i++){
14         memset(f[i&1],0x7f,sizeof(f[i&1]));
15         for(ll j=1;j<=n;j++){
16         f[i&1][j][w[i]]=min(f[i&1][j][w[i]],f[(i-1)&1][j][w[i-1]]+abs(w[i]-w[i-1]));//w[i-1]移动到w[i]
17         f[i&1][w[i]][w[i-1]]=min(f[i&1][w[i]][w[i-1]],f[(i-1)&1][j][w[i-1]]+abs(j-w[i]));//j移动到w[i]
18         f[i&1][w[i]][j]=min(f[i&1][w[i]][j],f[(i-1)&1][w[i-1]][j]+abs(w[i]-w[i-1]));//w[i-1]移动到w[i]
19         f[i&1][w[i-1]][w[i]]=min(f[i&1][w[i-1]][w[i]],f[(i-1)&1][w[i-1]][j]+abs(j-w[i]));//j移动到w[i]
20         }
21     }
22     ll ans=0x7fffffff;
23     for(ll j=1;j<=n;j++){
24         ans=min(ans,min(f[q&1][j][w[q]],f[q&1][w[q]][j]));
25     }
26     printf("%lld\n",ans);
27 }
View Code

30分算法

有一维一定是$w[i]$

考虑去掉一维

$f[i][j]=f[i-1][j]+abs(w[i]-w[i-1])$另一个指针从$w[i-1]$移动到$w[i]$

$f[i][w[i-1]]=f[i-1][j]+abs(j-w[i])$从$j$移动到$w[i]$

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll f[2][2019],w[2019];
ll a,b,n,q;
int main(){
    scanf("%lld%lld%lld%lld",&n,&q,&a,&b);
    memset(f,0x7f,sizeof(f));
    for(ll i=1;i<=q;i++)
        scanf("%lld",&w[i]);
    f[1][a]=abs(b-w[1]);
    f[1][b]=abs(a-w[1]);
    for(ll i=2;i<=q;i++){
        memset(f[i&1],0x7f,sizeof(f[i&1]));
        for(ll j=1;j<=n;j++){
        f[i&1][j]=min(f[i&1][j],f[(i-1)&1][j]+abs(w[i]-w[i-1]));
        f[i&1][w[i-1]]=min(f[i&1][w[i-1]],f[(i-1)&1][j]+abs(j-w[i]));
        f[i&1][j]=min(f[i&1][j],f[(i-1)&1][j]+abs(w[i]-w[i-1]));
        f[i&1][w[i-1]]=min(f[i&1][w[i-1]],f[(i-1)&1][j]+abs(j-w[i]));
        }
    }
    ll ans=0x7fffffff;
    for(ll j=1;j<=n;j++){
        ans=min(ans,min(f[q&1][j],f[q&1][j]));
    }
    printf("%lld\n",ans);
}
View Code

100分算法

两个转移式子

$f[i][j]=f[i-1][j]+abs(w[i]-w[i-1])$

$f[i][w[i-1]]=f[i-1][j]+abs(j-w[i])$

发现第一个式子就是区间加,第二个式子单点赋值

单点赋值赋的就是$min$,有个$abs$怎么办维护$f-j$最小值和$f+j$最小值

        memset(askmin,0x7f,sizeof(askmin));
        seg_min(1,1,w[i],2);//p[i]比当前点大,那么取p[i]-l
        seg_min(1,w[i],n,1);//p[i]比当前值小,取l-p[i]
    //    printf("ask=%lld %lld\n",askmin[1],askmin[2]);
        askmin[1]-=w[i];
        askmin[2]+=w[i];
        ll nowmin=min(askmin[1],askmin[2]);

线段树优化一下

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 ll w[1010101],askmin[4];
 5 ll a,b,n,q;
 6 struct tree{
 7     ll l,r,f,minn[4];
 8 }tr[1010101];
 9 void up(ll x){
10     for(ll i=1;i<=3;i++)
11         tr[x].minn[i]=min(tr[x<<1].minn[i],tr[x<<1|1].minn[i]);
12 }
13 void down(ll x){
14     tr[x<<1].f+=tr[x].f;
15     tr[x<<1|1].f+=tr[x].f;
16     for(ll i=1;i<=3;i++)
17         tr[x<<1].minn[i]+=tr[x].f,tr[x<<1|1].minn[i]+=tr[x].f;
18     tr[x].f=0;
19 }
20 void built(ll x,ll l,ll r){//printf("builx=%lld\n",x);
21     tr[x].l=l,tr[x].r=r;
22     if(l==r){
23         if(l==a||l==b){
24             tr[x].minn[3]=abs(a+b-l-w[1]);
25 //            printf("tr3=%lld\n",tr[x].minn[3]);
26             tr[x].minn[1]=tr[x].minn[3]+l;
27             tr[x].minn[2]=tr[x].minn[3]-l;
28         }
29         else tr[x].minn[1]=tr[x].minn[2]=tr[x].minn[3]=0x7ffffffffff;
30         return ;
31     }
32     ll mid=(l+r)>>1;
33     built(x<<1,l,mid);
34     built(x<<1|1,mid+1,r);
35     up(x);
36 }
37 void seg_min(ll x,ll l,ll r,ll zl){
38 //    printf("l=%lld r=%lld\n",l,r);
39     if(tr[x].l>=l&&tr[x].r<=r){
40 //        printf("tr[%lld].minn[%lld]=%lld l=%lld r=%lld\n",x,zl,tr[x].minn[zl],tr[x].l,tr[x].r);
41         askmin[zl]=min(askmin[zl],tr[x].minn[zl]);
42         return ;
43     }
44     down(x);
45     ll mid=(tr[x].l+tr[x].r)>>1;
46     if(mid>=l) seg_min(x<<1,l,r,zl);
47     if(mid<r) seg_min(x<<1|1,l,r,zl);
48     up(x);
49 }
50 void add(ll x,ll point,ll val){
51     if(tr[x].l==tr[x].r){
52         tr[x].minn[3]=min(tr[x].minn[3],val);
53         tr[x].minn[1]=tr[x].minn[3]+tr[x].l;
54         tr[x].minn[2]=tr[x].minn[3]-tr[x].l;
55         return ;
56     }
57     down(x);
58     ll mid=(tr[x].l+tr[x].r)>>1;
59     if(point<=mid) add(x<<1,point,val);
60     else add(x<<1|1,point,val);
61     up(x);
62 }
63 int main(){
64     scanf("%lld%lld%lld%lld",&n,&q,&a,&b);
65     for(ll i=1;i<=q;i++)
66         scanf("%lld",&w[i]);
67     built(1,1,n);
68     for(ll i=2;i<=q;i++){
69         memset(askmin,0x7f,sizeof(askmin));
70         seg_min(1,1,w[i],2);//p[i]比当前点大,那么取p[i]-l
71         seg_min(1,w[i],n,1);//p[i]比当前值小,取l-p[i]
72     //    printf("ask=%lld %lld\n",askmin[1],askmin[2]);
73         askmin[1]-=w[i];
74         askmin[2]+=w[i];
75         ll nowmin=min(askmin[1],askmin[2]);
76     //    printf("nowmin=%lld\n",nowmin);
77         tr[1].f+=abs(w[i]-w[i-1]);
78         for(ll j=1;j<=3;j++)
79             tr[1].minn[j]+=abs(w[i]-w[i-1]);
80         add(1,w[i-1],nowmin);
81     }
82     printf("%lld\n",tr[1].minn[3]);
83 }
View Code

 

posted @ 2019-09-17 16:36  znsbc  阅读(207)  评论(0)  编辑  收藏  举报