bzoj 1597

居然还有奶牛题没被权限(感动QAQ)

如果有包含关系的话就可以去掉小的，所以可以先排完序后去掉逆序的

然后长和宽都是单调的，就可以出方程f[i]=max{f[j]+a[j]b[i]}

易得(f[j-1]-f[k-1])/(a[j]-a[k])<b[i](易项时注意符号问题)

单调队列维护凸包即可

 1 #include<bits/stdc++.h>
 2 #define inc(i,l,r) for(int i=l;i<=r;i++)
 3 #define dec(i,l,r) for(int i=l;i>=r;i--)
 4 #define link(x) for(edge *j=h[x];j;j=j->next)
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define inf 1e9
 7 #define ll long long
 8 #define succ(x) (1<<x)
 9 #define NM 50000+5
10 using namespace std;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(isdigit(ch))x=x*10+ch-‘0‘,ch=getchar();
15     return x*f;
16 }
17 struct tmp{
18     int x,y;
19 }c[NM];
20 int n,q[NM],qh,qt,_t,a[NM],b[NM],m;
21 ll f[NM];
22 bool cmp(tmp x,tmp y){
23     return x.x>y.x||(x.x==y.x&&x.y>y.y);
24 }
25 double slope(int x,int y){
26     return (double)(f[y-1]-f[x-1])/(a[x]-a[y]);
27 }
28 int main(){
29     freopen("data.in","r",stdin);
30     n=read();
31     inc(i,1,n){
32         c[i].x=read();c[i].y=read();
33     }
34     sort(c+1,c+n+1,cmp);
35     inc(i,1,n)
36     if(_t<c[i].y){
37         a[++m]=c[i].x;b[m]=c[i].y;_t=c[i].y;
38     }
39 //    inc(i,1,n)printf("%d %d\n",a[i],b[i]);
40     qh=qt=1;q[1]=1;f[1]=(ll)a[1]*b[1];n=m;
41     inc(i,2,n){
42         while(qh+1<=qt&&slope(q[qh],q[qh+1])<b[i])qh++;
43         int j=q[qh];
44         f[i]=min(f[j-1]+(ll)a[j]*b[i],f[i-1]+(ll)a[i]*b[i]);
45         while(qh+1<=qt&&slope(q[qt-1],q[qt])>=slope(q[qt],i))qt--;
46         q[++qt]=i;
47     }
48 //    inc(i,1,n)printf("%d ",f[i]);printf("\n");
49     printf("%lld\n",f[n]);
50     return 0;
51 }