Dish too much.
T1 face the big board at a loss, then T2 greedy accidentally deleted sort statement. . .
T1 This large template ah. . . In fact, I think I can play out, and then an hour to code a 2k.
T2 then do want greedy came out. Ten minutes later the code finished T3 T1 fight to beat back violent instantaneous explosion.
Then again hit a 2k, WA0. To take discovery.
No cross-examination a few minutes and then walked violence.
Do not kick down the run, remember to take early improvement ideas.
T1:
Indeed quite bare tree line. Weights or discrete segment tree can be.
But the two are playing out on the test are dead.
Finally, the A discrete.
1 #include<cstdio> 2 #include<unordered_map> 3 #include<algorithm> 4 using namespace std; 5 #define inf 10000001 6 unordered_map<long long,int>M; 7 int m,cnt,k[100005];long long x[300005],l[100005],r[100005],re[300005]; 8 struct Segment_Tree{ 9 int cl[1500005],cr[1500005],lz[1500008],lzxor[1500008],w0[1500008],w1[1500008]; 10 void build(int p,int l,int r){ 11 cl[p]=l;cr[p]=r;lz[p]=-1;w0[p]=l;w1[p]=inf; 12 if(l==r)return; 13 build(p<<1,l,l+r>>1);build(p<<1|1,(l+r>>1)+1,r); 14 } 15 void up(int p){w0[p]=min(w0[p<<1],w0[p<<1|1]);w1[p]=min(w1[p<<1],w1[p<<1|1]);} 16 void down(int p){ 17 if(lz[p]!=-1){ 18 lz[p<<1]=lz[p],lz[p<<1|1]=lz[p]; 19 lzxor[p<<1]=lzxor[p<<1|1]=0; 20 if(lz[p])w1[p<<1]=cl[p<<1],w1[p<<1|1]=cl[p<<1|1],w0[p<<1]=w0[p<<1|1]=inf; 21 else w1[p<<1]=w1[p<<1|1]=inf,w0[p<<1]=cl[p<<1],w0[p<<1|1]=cl[p<<1|1]; 22 lz[p]=-1; 23 }else if(lzxor[p]){ 24 if(lz[p<<1]!=-1)lz[p<<1]^=1;else lzxor[p<<1]^=1; 25 if(lz[p<<1|1]!=-1)lz[p<<1|1]^=1;else lzxor[p<<1|1]^=1; 26 swap(w1[p<<1],w0[p<<1]); 27 swap(w1[p<<1|1],w0[p<<1|1]); 28 lzxor[p]=0; 29 } 30 up(p); 31 } 32 void set(int p,int l,int r,int w){ 33 if(l<=cl[p]&&cr[p]<=r){ 34 lz[p]=w;lzxor[p]=0; 35 if(w)w1[p]=cl[p],w0[p]=inf; 36 else w1[p]=inf,w0[p]=cl[p]; 37 return; 38 } 39 down(p); 40 if(l<=cr[p<<1])set(p<<1,l,r,w); 41 if(r>=cl[p<<1|1])set(p<<1|1,l,r,w); 42 up(p); 43 } 44 void Xor(int p,int l,int r){ 45 if(l<=cl[p]&&cr[p]<=r){ 46 if(lz[p]!=-1)lz[p]^=1;else lzxor[p]^=1; 47 swap(w0[p],w1[p]); 48 return; 49 } 50 down(p); 51 if(l<=cr[p<<1])Xor(p<<1,l,r); 52 if(r>=cl[p<<1|1])Xor(p<<1|1,l,r); 53 up(p); 54 } 55 }Tree; 56 main(){ 57 scanf("%d",&m); 58 for(int i=1;i<=m;++i)scanf("%d%lld%lld",&k[i],&l[i],&r[i]),x[i]=l[i],x[m+i]=r[i],x[m+m+i]=r[i]+1; 59 sort(x+1,x+1+m+m+m); 60 for(int i=1;i<=m*3;++i)if(x[i]!=x[i-1])M[x[i]]=++cnt,re[cnt]=x[i]; 61 for(int i=1;i<=m;++i)l[i]=M[l[i]],r[i]=M[r[i]]; 62 if(M.find(1)==M.end()){for(int i=1;i<=m;++i)puts("1");return 0;} 63 Tree.build(1,1,cnt);Tree.lz[1]=0; 64 for(int i=1;i<=m;++i){ 65 if(k[i]==1)Tree.set(1,l[i],r[i],1); 66 if(k[i]==2)Tree.set(1,l[i],r[i],0); 67 if(k[i]==3)Tree.Xor(1,l[i],r[i]); 68 printf("%lld\n",re[Tree.w0[1]]); 69 } 70 }
The accumulation of ideas:
- Segment tree template
T2:
In fact, one-third is not entirely correct. Although the secret nature proved that a single peak, but ooo given function value place outside the bottom is not strictly monotonic example.
Direct greedy, then we will find that decision a little complicated but may also go back.
But in fact, only four items, they look inside first row sequence (be sure to sort ah ah ah)
Follow the prompts range of data, both favorite items are special.
If we then determined the number of items to select both like, and the rest like that, greedy decisions.
The total cost was about it unimodal function (non-strict).
Note about the endpoint.
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 struct ps{ 5 int c1,c2;long long v; 6 friend bool operator<(ps a,ps b){return a.v<b.v;} 7 }p[100005]; 8 int n,m,k,a,b,n0,n1,n2,n3;long long v[100005],c1[100005],c2[100005]; 9 long long q0[100005],q1[100005],q2[100005],q3[100005],ans=100000000000000000ll; 10 long long chk(int p){ 11 long long tot=0,lft=m-k-(k-p); 12 for(int i=1;i<=p;++i)tot+=q3[i]; 13 for(int j=1;j<=k-p;++j)tot+=q1[j]+q2[j]; 14 int p0=1,p1=k-p+1,p2=k-p+1; 15 while(lft--) 16 if(q0[p0]<q1[p1]&&q0[p0]<q2[p2])tot+=q0[p0],p0++; 17 else if(q1[p1]<q2[p2])tot+=q1[p1],p1++; 18 else tot+=q2[p2],p2++; 19 ans=min(ans,tot);//printf("%d %lld\n",p,tot); 20 return tot; 21 } 22 main(){ 23 scanf("%d%d%d",&n,&m,&k); 24 for(int i=1;i<=n;++i)scanf("%lld",&v[i]); 25 scanf("%d",&a);for(int i=1,x;i<=a;++i)scanf("%d",&x),c1[x]=1; 26 scanf("%d",&b);for(int i=1,x;i<=b;++i)scanf("%d",&x),c2[x]=1; 27 for(int i=1;i<=n;++i) 28 if(c1[i]&&c2[i])q3[++n3]=v[i]; 29 else if(c1[i])q1[++n1]=v[i]; 30 else if(c2[i])q2[++n2]=v[i]; 31 else q0[++n0]=v[i]; 32 sort(q0+1,q0+n0+1); 33 sort(q1+1,q1+n1+1); 34 sort(q2+1,q2+n2+1); 35 sort(q3+1,q3+n3+1); 36 q0[n0+1]=q1[n1+1]=q2[n2+1]=1000000000000ll; 37 int l=0,r=n3; 38 l=max(l,max(k-n1,k-n2));l=max(l,2*k-m);//printf("%d %d\n",l,r); 39 if(l>r){puts("-1");return 0;} 40 while(l<r-1)if(chk(l+r>>1)<chk((l+r>>1)+1))r=l+r>>1;else l=l+r>>1; 41 for(int i=l;i<=r;++i)chk(i); 42 printf("%lld\n",ans); 43 }
- greedy
- Unimodal function thirds
- The two appeared together knowledge point total?
T3:
See below hair problem solution.
Very fairy.
1 #include<cstdio> 2 #include<bitset> 3 using namespace std; 4 bitset<401>B[401]; 5 int n,m,a[50005],b[50005],ans; 6 int main(){ 7 scanf("%d%d",&n,&m); 8 for(int i=1;i<=m;++i)scanf("%d%d",&a[i],&b[i]); 9 for(int i=1;i<=n;++i)B[i][i]=1; 10 for(int i=1;i<=n;++i)for(int j=m;j;--j) 11 if(B[i][a[j]]&&B[i][b[j]]){B[i].reset();break;} 12 else if(B[i][a[j]]&&!B[i][b[j]])B[i][b[j]]=1; 13 else if(B[i][b[j]]&&!B[i][a[j]])B[i][a[j]]=1; 14 for(int i=1;i<=n;++i)if(B[i].any())for(int j=i+1;j<=n;++j)if(B[j].any()&&(B[i]&B[j]).none())ans++; 15 printf("%d\n",ans); 16 }
When will return to its original state ah. . .
Why would such a dish ah. . .
But I would like to do ah. . .