pku1036Gangsters Dp

DP不小心写错还真的挺难调的= =

递推，时间复杂度（T*m+n）

状态为第i时间门的程度为j的最大繁荣值

Problem: 1036 User: BPM136
Memory: 704K Time: 125MS
Language: G++ Result: Accepted

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define LL long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
inline LL read()
{
LL d=0,f=1;char s=getchar();
while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
return d*f;
}
#define N 105
int f[2]
;
struct pack
{
int ti,w,c;
}a
;
int n,m,T;

bool acmp(pack a,pack b)
{
if(a.ti==b.ti)return a.w<b.w;
else return a.ti<b.ti;
}

int main()
{
n=read(),m=read(),T=read();
fo(i,0,n-1)a[i].ti=read();
fo(i,0,n-1)a[i].c=read();
fo(i,0,n-1)a[i].w=read();
sort(a,a+n,acmp);
memset(f,0,sizeof(f));int cnt=0;
fo(i,0,T)
{
fo(j,0,min(m,i))
{
int t=max(f[(i+1)%2][max(j-1,0)],f[(i+1)%2][min(j+1,m)]);
f[i%2][j]=max(t,f[(i+1)%2][j]);
}
while(a[cnt].ti==i&&cnt<n)
{
if(a[cnt].w<=i)f[i%2][a[cnt].w]+=a[cnt].c;
cnt++;
}
// cout<<"Case:"<<i<<' ';fo(j,1,m)cout<<f[i%2][j]<<' ';cout<<endl;
}
int ans=0;
fo(i,0,m)ans=max(ans,f[T%2][i]);
cout<<ans<<endl;
return 0;
}