T3 linked write segment tree, depressed ~! ! ! ~
prob1: Nakano (buff)
Subject to the effect: the same 2019.7.23 \ (T3 \) feast ( \ (Party \) )
\ (WoCao \) , wrote the original hang, do not want to say what
The original title will not talk about ideas, and directly on the code (later to pay attention to the overall classification):
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define int long long
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
const int xx=4010;
int dis[xx][xx],vis[xx],que[xx];
int head[xx],nxt[xx<<1],to[xx<<1],cnt=0;
inline void add(int u,int v){nxt[++cnt]=head[u];to[cnt]=v;head[u]=cnt;}
inline void bfs(int g)
{
dis[g][g]=0;
vis[g]=g;
int hd=0,tl=-1;
que[++tl]=g;
while(hd<=tl)
{
int gg=que[hd++];
for(int j=head[gg];j;j=nxt[j])
{
int jj=to[j];
if(vis[jj]!=g)
{
vis[jj]=g;
dis[g][jj]=dis[g][gg]+1;
que[++tl]=jj;
}
}
}
}
signed main()
{
int n=in,m=in;
fur(i,1,m)
{
int x=in,y=in;
add(x,y);
add(y,x);
}
memset(dis,0x7f,sizeof(dis));
fur(i,1,n) bfs(i);
int s1=in,t1=in,p1=in;
int s2=in,t2=in,p2=in;
if(dis[s1][t1]>p1||dis[s2][t2]>p2)
{
printf("-1\n");
return 0;
}
int ans=dis[s1][t1]+dis[s2][t2];
fur(i,1,n) fur(j,1,n)
{
if(dis[s1][i]+dis[i][j]+dis[j][t1]<=p1&&dis[s2][i]+dis[i][j]+dis[j][t2]<=p2) ans=min(ans,dis[s1][i]+dis[s2][i]+dis[i][j]+dis[j][t1]+dis[j][t2]);
if(dis[s1][j]+dis[j][i]+dis[i][t1]<=p1&&dis[s2][i]+dis[i][j]+dis[j][t2]<=p2) ans=min(ans,dis[s1][j]+dis[s2][i]+dis[j][i]+dis[i][t1]+dis[j][t2]);
}
printf("%lld\n",m-ans);
return 0;
}prob3: Genting Yi (cloud)
Title effect: known \ (n-\) th tuple \ ((a_i, B_i) \) , and the value \ (W \) , there are two variables \ (P, Q \) , when there are \ (m_1 \) th tuple satisfies \ (b_i <= q \) , the answers have \ (m_1 * q * w \ ) contribution, and if there are other tuple \ (M_2 \) bins and satisfying \ (a_i <= P \) , the answer will have (m_2 * p \) \ contributions when seeking \ (Q \) are equal to \ (1 \) to \ (max (b_i) +1 \ ) , the the maximum value of the answer
Strong sub-divided portion 50 : a maintenance and tub exhaust tuple list, the answer with the range segment tree maintenance, \ (O (logN-n-) \) modified:
#include<bits/stdc++.h>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define int long long
inline int read()
{
int x=0;
char ch=getchar();
for(;!isalnum(ch);ch=getchar());
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x;
}
const int xx=1e5+10;
struct linetree
{
int sz,ans;
linetree *son[2];
linetree():sz(0),ans(0)
{
son[0]=son[1]=NULL;
}
};
linetree *root,*ty,pool[xx<<2];
int res[xx],n,w,head[xx],nxt[xx],a[xx],wuhan[xx],mx=0,alll;
inline linetree *newlt()
{
*++ty=linetree();
return ty;
}
inline void up(linetree *k)
{
k->ans=0;
if(k->son[0]) k->ans=max(k->ans,k->son[0]->ans);
if(k->son[1]) k->ans=max(k->ans,k->son[1]->ans);
}
inline void add(linetree *k,int l,int r,int c)
{
++k->sz;
if(l==r)
{
k->ans=k->sz*wuhan[l];
return;
}
int mid=(l+r)>>1;
if(!k->son[0]) k->son[0]=newlt();
add(k->son[0],l,mid,c);
if(c>mid)
{
if(!k->son[1]) k->son[1]=newlt();
add(k->son[1],mid+1,r,c);
up(k);
}
else k->ans=max(k->son[0]->ans,k->ans);
}
inline void ade(int u,int v)
{
nxt[v]=head[u];
head[u]=v;
}
inline void sol()
{
int all=n;
fur(c,1,mx)
{
for(int j=head[c-1];j;j=nxt[j])
{
add(root,1,alll,a[j]);
--all;
}
res[c]=all*c*w+root->ans;
}
}
signed main()
{
n=in;w=in;
fur(i,1,n)
{
wuhan[i]=a[i]=in;
int y=in;
ade(y,i);
mx=max(mx,y);
}
sort(wuhan+1,wuhan+n+1);
alll=unique(wuhan+1,wuhan+n+1)-wuhan-1;
fur(i,1,n) a[i]=lower_bound(wuhan+1,wuhan+alll+1,a[i])-wuhan;
++mx;
ty=pool;
root=newlt();
sol();
fur(i,1,mx) printf("%lld ",res[i]);printf("\n");
return 0;
}
Positive Solutions : Block + optimized slope (push themselves persimmon)
#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define fdr(i,a,b) for(int i=a;i>=b;--i)
#define int long long
#define jiba signed
#define K(j,i) (double)(d[i]-d[j])/(j-i)
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
const int xx=1e5+10;
const int yy=1e3+10;
int pos[xx],d[xx],cnt[xx],tp=0,mx=0;
int que[yy][yy],ls[yy],rs[yy],tag[yy],ans[yy],hd[yy],tl[yy];
int res[xx],n,w,a[xx],head[xx],nxt[xx];
inline int get(int k,int i){return tag[k]*i+d[i];}
inline void add(int u,int to){nxt[to]=head[u];head[u]=to;}
inline void build(int k)
{
d[rs[k]]=rs[k]*cnt[rs[k]];
fdr(i,rs[k]-1,ls[k]) d[i]=(d[i+1]/(i+1)+cnt[i])*i;
hd[k]=1,tl[k]=1;
fur(i,ls[k],rs[k])
{
while(K(i,que[k][tl[k]])<K(que[k][tl[k]],que[k][tl[k]-1])&&tl[k]>hd[k]) --tl[k];
que[k][++tl[k]]=i;
}
while(get(k,que[k][hd[k]])<get(k,que[k][hd[k]+1])&&hd[k]<tl[k]) hd[k]++;
ans[k]=get(k,que[k][hd[k]]);
}
inline void ad_tag(int k)
{
while(get(k,que[k][hd[k]])<get(k,que[k][hd[k]+1])&&hd[k]<tl[k]) hd[k]++;
ans[k]=get(k,que[k][hd[k]]);
}
inline void sol()
{
int all=n;
fur(i,1,tp)
{
for(int j=head[i-1];j;j=nxt[j])
{
--all;
++cnt[a[j]];
fur(k,1,pos[a[j]]-1) ++tag[k],ad_tag(k);
build(pos[a[j]]);
}
fur(j,1,pos[mx]) res[i]=max(res[i],ans[j]);
res[i]+=all*w*i;
}
}
jiba main()
{
n=in;w=in;
fur(i,1,n)
{
a[i]=in;
mx=max(mx,a[i]);
int y=in;
tp=max(tp,y);
add(y,i);
}
int tmp=sqrt(mx);
fur(i,1,mx)
{
pos[i]=(i-1)/tmp+1;
if(pos[i]!=pos[i-1]) ls[pos[i]]=i;
rs[pos[i]]=i;
}
tp++;
sol();
fur(i,1,tp) printf("%lld ",res[i]);printf("\n");
return 0;
}
prob2: run fast ( \ (PDK \) )
Title effect: doudizhu (change), can be singles, pairs, three with two bombs (not the number of steps), straight, double straight, three straight, find the smallest step
Positive Solutions : \ (the DP \) + partial search, \ (the DP \) pretreated case singles, pairs, two of the three band, burst search various straight
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define fdr(i,a,b) for(int i=a;i>=b;--i)
#define int long long
#define jiba signed
inline int read()
{
int x=0,f=1;char ch=getchar();
for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*f;
}
int n,ans;
int f[7][9][13][25],cnt[14],upstair[14]={0,4,4,4,4,4,4,4,4,4,4,4,3,1};
inline void up(int &a,int b){a=a>b?b:a;}
inline void init()
{
fur(i,0,n/4+1) fur(j,0,n/3+1) fur(k,0,n/2+1) fur(l,0,n+1) if(i*4+j*3+k*2+l<=n)
{
if(i)
{
up(f[i][j][k][l],f[i-1][j][k][l]);
up(f[i][j][k][l],f[i-1][j+1][k][l+1]);
up(f[i][j][k][l],f[i-1][j][k+2][l]);
up(f[i][j][k][l],f[i-1][j][k][l+4]);
}
if(j)
{
up(f[i][j][k][l],f[i][j-1][k][l]+1);
if(k) up(f[i][j][k][l],f[i][j-1][k-1][l]+1);
if(l) up(f[i][j][k][l],f[i][j-1][k][l-1]+1);
if(l>=2) up(f[i][j][k][l],f[i][j-1][k][l-2]+1);
up(f[i][j][k][l],f[i][j-1][k+1][l+1]);
up(f[i][j][k][l],f[i][j-1][k][l+3]);
}
if(k)
{
up(f[i][j][k][l],f[i][j][k-1][l]+1);
up(f[i][j][k][l],f[i][j][k-1][l+2]);
}
if(l) up(f[i][j][k][l],f[i][j][k][l-1]+1);
}
}
inline void search(int w,int las)
{
if(w>=ans) return;
if(las==0) return;
int rest[5]={0,0,0,0,0};
fur(i,1,13) if(cnt[i]) rest[cnt[i]]++;
up(ans,w+f[rest[4]][rest[3]][rest[2]][rest[1]]);
fur(i,1,11)
{
if(cnt[i]>=3)
{
cnt[i]-=3;
int j=i+1;
while(j<=12&&cnt[j]>=3)
{
cnt[j]-=3;
search(w+1,las-(j-i+1)*3);
++j;
}
fur(q,i,j-1) cnt[q]+=3;
}
}
fur(i,1,10)
{
if(cnt[i]>=2&&cnt[i+1]>=2)
{
cnt[i]-=2;
cnt[i+1]-=2;
int j=i+2;
while(j<=12&&cnt[j]>=2)
{
cnt[j]-=2;
search(w+1,las-(j-i+1)*2);
++j;
}
fur(q,i,j-1) cnt[q]+=2;
}
}
fur(i,1,8)
{
if(cnt[i]&&cnt[i+1]&&cnt[i+2]&&cnt[i+3])
{
cnt[i]-=1;
cnt[i+1]-=1;
cnt[i+2]-=1;
cnt[i+3]-=1;
int j=i+4;
while(j<=12&&cnt[j])
{
cnt[j]-=1;
search(w+1,las-(j-i+1));
++j;
}
fur(q,i,j-1) cnt[q]+=1;
}
}
}
jiba main()
{
freopen("pdk.in","r",stdin);
freopen("pdk.out","w",stdout);
int t=in;n=in+1;
memset(f,0x3f,sizeof(f));
f[0][0][0][1]=0;
init();
while(t--)
{
memset(cnt,0,sizeof(cnt));
ans=23;
fur(i,1,n-1)
{
int x=in,y=in;
if(x==1) cnt[12]++;
else if(x==2) cnt[13]++;
else cnt[x-2]++;
}
fur(i,1,12)
{
if(cnt[i]<upstair[i])
{
++cnt[i];
search(0,n);
--cnt[i];
}
}
printf("%lld\n",ans);
}
return 0;
}