Bzoj3163

多重背包单调队列优化正反跑一遍 $O(nm)$

查询时,合并 $f_{d-1,e}+g_{d+1,e}$。

不一定要全取完,所以做一个前缀最大。

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#include<bits/stdc++.h>
#define maxn 1005
#define ll long long
#define put() putchar('\n')
using namespace std;
inline void read(int &x){
int f=1;x=0;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
x*=f;
}
int n;
struct yyy{
int v,w,k;
}a[1005];
int f[1005][1005],g[1005][1005];
int head,tail,q[1005],m=1000;
int suf[maxn],suff[maxn];
signed main(void){
freopen("1.in","r",stdin);
int i,j,k,T,x,y;
read(n);
for (i=1;i<=n;i++) read(a[i].v),read(a[i].w),read(a[i].k);
for (i=1;i<=n;i++) {
for (j=0;j<a[i].v;j++) {
q[head=tail=1]=0;
f[i][j]=f[i-1][j];
for (k=1;k*a[i].v+j<=m;k++) {
while (head<=tail&&q[head]+a[i].k<k) head++;
while (head<=tail&&f[i-1][j+q[tail]*a[i].v]-q[tail]*a[i].w<=f[i-1][j+k*a[i].v]-k*a[i].w) tail--;
q[++tail]=k;
f[i][j+k*a[i].v]=f[i-1][j+q[head]*a[i].v]+(k-q[head])*a[i].w;
}
}
// for (j=0;j<=8;j++) printf("%d ",f[i][j]);put();
}
for (i=n;i>=1;i--) {
for (j=0;j<a[i].v;j++) {
q[head=tail=1]=0;
g[i][j]=g[i+1][j];
for (k=1;k*a[i].v+j<=m;k++) {
while (head<=tail&&q[head]+a[i].k<k) head++;
while (head<=tail&&g[i+1][j+q[tail]*a[i].v]-q[tail]*a[i].w<=g[i+1][j+k*a[i].v]-k*a[i].w) tail--;
q[++tail]=k;
g[i][j+k*a[i].v]=g[i+1][j+q[head]*a[i].v]+(k-q[head])*a[i].w;
}
}
}
read(T);
while (T--){
read(x);read(y);x++;
int ans=0;
for (i=0;i<=y;i++) suf[i]=max(suf[i-1],f[x-1][i]),suff[i]=max(suff[i-1],g[x+1][i]);
for (i=0;i<=y;i++) ans=max(ans,suf[i]+suff[y-i]);
printf("%d\n",ans);
}
return 0;
}