And the burst zero! ! T_T ......
prob1: Flush ( \ (Card \) )
Sucker question did not even write it, depressed ......
Consider cards \ (I \) is the number of cards with the end of the flush, then certainly will change to \ (num = \ sum [b_j <b_i-n + 1] + \ sum [a_j \ ne a_i \ land b_j> B_i. 1-n-+ = \ Land b_j <= B_i] + \ SUM [b_j> B_i] \) .
Well, I see persimmon not want to do ......
But in fact, if a complement thought wave can be considered into \ (NUM = N- \ SUM [a_j = a_i \ Land b_j> = B_i. 1-n-+ \ Land b_j <= B_i] \) .
Persimmon so there is a lot of it ......
Another color for the first keyword, second grade sort keyword, you can do a very good cycle pruning \ (BREAK \) , removed a lot of unnecessary operations, the complexity of the \ (O ( n ^ 2) \) becomes \ (O \) (can live), very good.
Finally, make up one: remember to heavy.
#include<iostream>
#include<cstdio>
#include<algorithm>
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 jinitaimei 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;
}
const int xx=1e5+10;
struct prom{int c,n;}b[xx],a[xx];
inline bool cmp(prom x,prom y){return (x.c^y.c)?(x.c<y.c):(x.n<y.n);}
jinitaimei main()
{
int n=in,cnt=0;
fur(i,1,n) a[i]=(prom){in,in};
sort(a+1,a+n+1,cmp);
fur(i,1,n)
{
while(a[i].c==a[i+1].c&&a[i].n==a[i+1].n) ++i;
b[++cnt]=a[i];
}
int ans=0;
fur(i,1,cnt)
{
int tmp=0;
fdr(j,i,1)
{
if(b[j].c==a[i].c&&b[j].n>=a[i].n-n+1) ++tmp;
else break;
}
ans=max(ans,tmp);
}
printf("%lld\n",n-ans);
return 0;
}
prob2: an experiment ( \ (the Test \) )
Enumerator \ (a \) of each subset, asking whether the time its last appearance in the \ (ib \) and \ (i-1 \) between, and if so, there is some contribution, while updating the subset The last time appears.
\ (30 \) in portions minutes, or less violent enumerate all \ (A \) number \ (I \) , it is determined whether \ (A \) subset, and accumulating contributions:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;++i)
#define int long long
#define jinitaimei 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;
}
const int xx=1e5+10;
int f[xx];
jinitaimei main()
{
int n=in;
fur(i,1,n)
{
int x=in,y=in,res=0;
fur(j,1,x)
{
if((j|x)!=x) continue;
if(f[j]<i-y) ++res;
f[j]=i;
}
printf("%lld\n",res);
}
return 0;
}
Positive Solution : Use Bitwise & enumerate the subset (push themselves just fine):
#include<bits/stdc++.h>
using namespace std;
#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 in read()
#define int long long
#define jinitaimei signed
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;
int las[xx];
jinitaimei main()
{
int n=in;
fur(i,1,n)
{
int a=in,b=in,res=0;
for(int j=a;j;j=(j-1)&a)
{
if(las[j]<i-b) ++res;
las[j]=i;
}
printf("%lld\n",res);
}
return 0;
}
It looks like a standard code is shorter than violence? ?
prob3: save the world ( \ (the Save \) )
Clear thinking: \ (Tarjan \) condensing point + (SPFA \) \ longest road, taking the end of the last enumerated \ (max \) can.
(Looks like \ (Tarjan \) write ugly):
#include<bits/stdc++.h>
using namespace std;
#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 jinitaimei signed
#define pi pair<int,int>
#define nm make_pair
#define in read()
inline int read()
{
int x=0,k=1;
char ch=getchar();
for(;!isalnum(ch);ch=getchar()) k=(ch=='-'?-1:1);
for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
return x*k;
}
const int xx=5e5+10;
int sta[xx],top=0,all=0;
int low[xx],id[xx],cnt=0;
int sum[xx],w[xx];
int bel[xx],bar[xx];
int u[xx],v[xx],dis[xx];
bool vis[xx],viss[xx];
vector<int>e[xx];
vector<pi>f[xx];
deque<int>p;
inline void Tarjan(int g)
{
low[g]=id[g]=++cnt;
sta[++top]=g;
vis[g]=true;
fur(i,0,(int)e[g].size()-1)
{
if(!id[e[g][i]])
{
Tarjan(e[g][i]);
low[g]=min(low[g],low[e[g][i]]);
}
else if(vis[e[g][i]]) low[g]=min(low[g],id[e[g][i]]);
}
if(id[g]==low[g])
{
++all;
do
{
int tp=sta[top];
sum[all]+=w[tp];
vis[tp]=false;
bel[tp]=all;
}while(sta[top--]!=g);
}
}
inline void spfa(int g)
{
fur(i,1,all) dis[i]=0,viss[i]=false;
int s=bel[g];
p.push_back(s);
dis[s]=sum[s];
viss[s]=true;
while(!p.empty())
{
int hd=p.front();
p.pop_front();
viss[hd]=false;
fur(i,0,(int)f[hd].size()-1)
{
if(f[hd][i].second+dis[hd]>dis[f[hd][i].first])
{
dis[f[hd][i].first]=dis[hd]+f[hd][i].second;
if(!viss[f[hd][i].first])
{
viss[f[hd][i].first]=true;
if(!p.empty())
{
if(dis[f[hd][i].first]>dis[p.front()]) p.push_front(f[hd][i].first);
else p.push_back(f[hd][i].first);
}
else p.push_back(f[hd][i].first);
}
}
}
}
}
jinitaimei main()
{
int n=in,m=in;
fur(i,1,m)
{
u[i]=in,v[i]=in;
if(u[i]==v[i]) continue;
e[u[i]].push_back(v[i]);
}
fur(i,1,n) w[i]=in;
fur(i,1,n) if(!id[i]) Tarjan(i);
fur(i,1,m) if(bel[u[i]]!=bel[v[i]]) f[bel[u[i]]].push_back(nm(bel[v[i]],sum[bel[v[i]]]));
int s=in,p=in;
fur(i,1,p) bar[i]=in;
spfa(s);
int ans=0;
fur(i,1,p) ans=max(ans,dis[bel[bar[i]]]);
printf("%lld\n",ans);
return 0;
}