LINK:序列
考虑前20分 容易想到爆搜。
考虑dp 容易设\(f_{i,j,k,l}\)表示前i个位置 选了j对 且此时A选择了k个 B选择了l个的最大值.期望得分28.
code
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 1000000000
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d\n",x)
#define putl(x) printf("%lld\n",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-4
#define sq sqrt
#define F first
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
RE int x=0,f=1;RE char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
return x*f;
}
const int MAXN=32;
int T,n,K,L;
int a[MAXN],b[MAXN];
ll f[MAXN][MAXN][MAXN][MAXN];
int main()
{
//freopen("1.in","r",stdin);
get(T);
while(T--)
{
memset(f,0xcf,sizeof(f));
get(n);get(K);get(L);
rep(1,n,i)get(a[i]);
rep(1,n,i)get(b[i]);
f[0][0][0][0]=0;
rep(1,n,i)
{
int cc=min(i-1,L);
rep(0,cc,j)
{
int ww=min(i-1,K);
rep(0,ww,l)rep(0,ww,r)
{
if(l+1<=K&&r+1<=K&&j+1<=L)f[i][j+1][l+1][r+1]=max(f[i-1][j][l][r]+a[i]+b[i],f[i][j+1][l+1][r+1]);
if(l+1<=K&&r+1<=K)f[i][j][l+1][r+1]=max(f[i-1][j][l][r]+a[i]+b[i],f[i][j][l+1][r+1]);
if(l+1<=K)f[i][j][l+1][r]=max(f[i-1][j][l][r]+a[i],f[i][j][l+1][r]);
if(r+1<=K)f[i][j][l][r+1]=max(f[i-1][j][l][r]+b[i],f[i][j][l][r+1]);
f[i][j][l][r]=max(f[i][j][l][r],f[i-1][j][l][r]);
}
}
}
putl(f[n][L][K][K]);
}
return 0;
}
容易发现这可以贪心 考虑费用流.遗憾的是 考场上我建了2h的图没建出来。
最近 我又建了1h图 还是没有什么进展 看完题解的建图恍然大悟.
我一直考虑把限制少的先建立出来 把限制多的放后边加边来进行限制。
其实并非如此.先考虑 将限制多的先建立出来.
即先限流L 然后流向每个i点流量费用\((1,A[i])\) 然后 i+n\((1,B[i])\) 然后 i向i+n连边。
接下来处理那些随便流的。i连向一个公共点cc 然限流 cc-zz (K-L,0) zz连向i+n的那些点即可.
期望的分64?实际得分48.
code
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 10000000000000000ll
#define inf 1000000000
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d\n",x)
#define putl(x) printf("%lld\n",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-4
#define sq sqrt
#define S second
#define F first
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
RE int x=0,f=1;RE char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
return x*f;
}
const int MAXN=200010,N=MAXN<<1,maxn=MAXN*5<<1;
int n,m,T,TT,zz,cc,S,SS,K,L,len;
int a[MAXN],b[MAXN],in[N],vis[N],pre[N];ll dis[N],ans;
int lin[N],ver[maxn],e[maxn],e1[maxn],nex[maxn],q[maxn<<1];
inline void add(int x,int y,int z,int z1)
{
ver[++len]=y;nex[len]=lin[x];lin[x]=len;e[len]=z;e1[len]=z1;
ver[++len]=x;nex[len]=lin[y];lin[y]=len;e[len]=0;e1[len]=-z1;
}
inline bool spfa()
{
rep(1,T,i)dis[i]=-INF;
dis[S]=0;in[S]=inf;int l=0,r;
q[r=1]=S;vis[S]=1;
//cout<<l<<endl;
while(++l<=r)
{
int x=q[l];vis[x]=0;
go(x)
{
if(!e[i])continue;
//cout<<tn<<endl;
if(dis[tn]<dis[x]+e1[i])
{
dis[tn]=dis[x]+e1[i];
if(!vis[tn])q[++r]=tn,vis[tn]=1;
in[tn]=min(in[x],e[i]);
pre[tn]=i;
}
}
}
//cout<<dis[T]<<endl;
return dis[T]!=-INF;
}
inline void EK()
{
ans=0;
//cout<<1<<endl;
while(spfa())
{
int x=T,i=pre[x];
ans+=dis[T]*in[T];
while(x!=S)
{
e[i]-=in[T];
e[i^1]+=in[T];
x=ver[i^1];i=pre[x];
}
}
}
int main()
{
//freopen("1.in","r",stdin);
get(TT);
while(TT--)
{
rep(1,T,i)lin[i]=0;len=1;
get(n);get(K);get(L);
rep(1,n,i)get(a[i]);
rep(1,n,i)get(b[i]);
zz=n<<1|1;cc=zz+1;S=cc+1;SS=S+1;T=SS+1;
add(S,SS,K,0);
add(zz,cc,K-L,0);
rep(1,n,i)
{
add(i,i+n,1,0);
add(i,zz,1,0);
add(SS,i,1,a[i]);
add(cc,i+n,1,0);
add(i+n,T,1,b[i]);
}
//cout<<len<<endl;
EK();putl(ans);
}
return 0;
}
考虑正解 费用流是正确的 不过复杂度太高了 没有保证.
我们可以进行模拟费用流来将复杂度降下来。
所以操作都模拟费用流时 优先跑增广路的权值最大的即可.
其中有些细节：因为退流的操作很难模拟出来,所以我们考虑当退流的时候 其实是选择一个和其匹配的点.
综上 我们这个模拟费用流不退流 每个被选择的数字都是最终的答案的一部分.
操作：显然应该先选出K-L的自由流.
然后考虑三种情况 一种是a[x]+b[x] 一种是给已经选了A[x]的但没有选B[x]的配一个B[x]以及剩下的A中的最大值.
还有一种和前者相反 讨论清楚 开堆模拟即可.
容易证明复杂度为\(nlogn\)
code
//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 10000000000000000ll
#define inf 1000000000
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d\n",x)
#define putl(x) printf("%lld\n",x)
#define rep(p,n,i) for(RE int i=p;i<=n;++i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE int i=n;i>=p;--i)
#define vep(p,n,i) for(RE int i=p;i<n;++i)
#define pii pair<int,int>
#define mk make_pair
#define RE register
#define P 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-4
#define sq sqrt
#define S second
#define F first
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline int read()
{
RE int x=0,f=1;RE char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();}
return x*f;
}
const int MAXN=200010,N=MAXN<<1,maxn=MAXN*5<<1;
int n,m,K,L,T;ll ans=0;
int a[MAXN],b[MAXN],id[MAXN],s[MAXN];
struct A{int id;inline bool friend operator <(A x,A y){return a[x.id]<a[y.id];}}w1;
struct B{int id;inline bool friend operator <(B x,B y){return b[x.id]<b[y.id];}}w2;
struct AB{int id;inline bool friend operator <(AB x,AB y){return a[x.id]+b[x.id]<b[y.id]+a[y.id];}}w3;
priority_queue<A>H1,F1;
priority_queue<B>H2,F2;
priority_queue<AB>H3;
inline int cmpa(int x,int y){return a[x]>a[y];}
inline int cmpb(int x,int y){return b[x]>b[y];}
inline void solve(int m)
{
rep(1,n,i)id[i]=i;ans=0;
sort(id+1,id+1+n,cmpa);rep(1,m,i)ans+=a[id[i]],s[id[i]]=1;
sort(id+1,id+1+n,cmpb);rep(1,m,i)ans+=b[id[i]],s[id[i]]|=2;
}
int main()
{
//freopen("1.in","r",stdin);
get(T);
while(T--)
{
get(n);get(K);get(L);
rep(1,n,i)get(a[i]),s[i]=0;
rep(1,n,i)get(b[i]);
while(H1.size())H1.pop();
while(H2.size())H2.pop();
while(H3.size())H3.pop();
while(F1.size())F1.pop();
while(F2.size())F2.pop();
solve(K-L);
int res=0;
rep(1,n,i)
{
if(s[i]==3){++res;continue;}
if(!s[i])
{
w1.id=i;H1.push(w1);
w2.id=i;H2.push(w2);
w3.id=i;H3.push(w3);
}
if(s[i]==1)w2.id=i,H2.push(w2),F2.push(w2);
if(s[i]==2)w1.id=i,H1.push(w1),F1.push(w1);
}
int id1,id2,v1,v2,v3,c1,c2,mx;
while(L--)
{
while(H1.size()&&(s[H1.top().id]&1))H1.pop();
while(F1.size()&&(s[F1.top().id]&1))F1.pop();
while(H2.size()&&(s[H2.top().id]&2))H2.pop();
while(F2.size()&&(s[F2.top().id]&2))F2.pop();
while(H3.size()&&s[H3.top().id])H3.pop();
if(res)
{
--res;
id1=H1.top().id,id2=H2.top().id;
ans+=a[id1];ans+=b[id2];
s[id1]|=1;s[id2]|=2;
if(id1==id2){++res;continue;}
if(s[id1]==3)++res;
if(s[id2]==3)++res;
if(s[id1]==1)w2.id=id1,F2.push(w2);
if(s[id2]==2)w1.id=id2,F1.push(w1);
continue;
}
v1=0,v2=0,v3=0;
if(F1.size()){id2=H2.top().id;v1=a[F1.top().id]+b[id2],c1=id2==1?1:0;}
if(F2.size()){id1=H1.top().id;v2=a[id1]+b[F2.top().id],c2=id1==2?1:0;}
if(H3.size()){id1=H3.top().id;v3=a[id1]+b[id1];}
mx=max(max(v1,v2),v3);
if(v1==v2&&v2==mx)
{
if(c1>=c2)
{
id2=H2.top().id;id1=F1.top().id;
ans+=a[id1]+b[id2];
s[id1]|=1;s[id2]|=2;
if(s[id2]==3)++res;
else w1.id=id2,F1.push(w1);
}
else
{
id1=H1.top().id;id2=F2.top().id;
ans+=a[id1]+b[id2];
s[id1]|=1;s[id2]|=2;
if(s[id1]==3)++res;
else w2.id=id1,F2.push(w2);
}
continue;
}
if(v1==mx)
{
id2=H2.top().id;id1=F1.top().id;
ans+=a[id1]+b[id2];
s[id1]|=1;s[id2]|=2;
if(s[id2]==3)++res;
else w1.id=id2,F1.push(w1);
continue;
}
if(v2==mx)
{
id1=H1.top().id;id2=F2.top().id;
ans+=a[id1]+b[id2];
s[id1]|=1;s[id2]|=2;
if(s[id1]==3)++res;
else w2.id=id1,F2.push(w2);
continue;
}
if(mx==v3)
{
id1=H3.top().id;
s[id1]=3;ans+=mx;
continue;
}
}
putl(ans);
}
return 0;
}