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 }
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 }
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 }
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 }
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); }
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]-lseg_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 }
