P2120 [ZJOI2007]仓库建设

P2120 [ZJOI2007]仓库建设

怎么说呢？算是很水的题了吧...

只要不要一开始就把dp想错就行...

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e6+10;
const ll INF=2e18;
ll n,f[N],x[N],p[N],c[N],q[N],l,r,sum[N],sump[N];//f[i][0/1]表示i及之后的工厂产品
inline int read()//i工厂建仓库的最小代价
{
    int x=0,ff=1;
    char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') ff=-1;ch=getchar();}
    while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*ff;
}
inline double X(int i) {return sump[i];}
inline double Y(int i) {return f[i]+sum[i];}
inline double xie(int i,int j) {return (Y(i)-Y(j))/(X(i)-X(j));}
int main()
{
//    freopen("1.in","r",stdin);
    n=read();
    for(int i=1;i<=n;++i) x[i]=read(),p[i]=read(),c[i]=read();
    for(int i=1;i<=n;++i) sum[i]=sum[i-1]+x[i]*p[i],sump[i]=sump[i-1]+p[i];
    for(int i=1;i<=n;++i) f[i]=INF;
    f[0]=0;
    for(int i=1;i<=n;++i)
    {
        while(l<r&&xie(q[l],q[l+1])<=x[i]) ++l;
        int j=q[l];
        f[i]=f[j]+(sump[i-1]-sump[j])*x[i]-(sum[i-1]-sum[j])+c[i];
        while(l<r&&xie(i,q[r])<=xie(q[r],q[r-1])) --r;
        q[++r]=i;
        //for(int j=0;j<i;++j) f[i]=min(f[i],f[j]+(sump[i-1]-sump[j])*x[i]-(sum[i-1]-sum[j]));
        //f[i]+=c[i];
    }
    printf("%lld",f[n]);
    return 0;
}