[CEOI2004]锯木厂选址

link

做这种题就应该去先写个暴力代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){ans=ans*10+c-‘0‘;c=getchar();}
    return f*ans;
}
const int N=200011;
int w[N],n,f[N],deep[N],sum,s[N],p[N],k[N];
int calc(int l,int r){
    return (p[r]-p[l-1])-deep[r]*(k[r]-k[l-1]);
}
int main(){
//    freopen("5.in","r",stdin);
    memset(f,127/3,sizeof(f));f[0]=0;
    n=read();
    for(int i=1;i<=n;i++) w[i]=read(),deep[i]=read();
    for(int i=n;i>=1;i--) deep[i]+=deep[i+1];
    n++;
    for(int i=1;i<=n;i++) s[i]=w[i]*deep[i];
    for(int i=1;i<=n;i++) p[i]=p[i-1]+s[i];
    for(int i=1;i<=n;i++) k[i]=k[i-1]+w[i];
    for(int i=1;i<=n;i++){
        for(int j=0;j<i;j++){
            f[i]=min(f[i],-deep[j]*k[j]-deep[i]*k[i]+deep[i]*k[j]+deep[n]*k[i]);
        }
    }
    int maxn=INT_MAX;
    for(int i=1;i<=n;i++) maxn=min(maxn,f[i]);cout<<maxn-deep[n]*k[n]+p[n];
}

1

然后再把calc放在里面,把无用的东西提出去。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){ans=ans*10+c-‘0‘;c=getchar();}
    return f*ans;
}
const int N=200011;
int w[N],n,f[N],deep[N],sum,s[N],p[N],k[N];
int calc(int l,int r){
    return (p[r]-p[l-1])-deep[r]*(k[r]-k[l-1]);
}
int main(){
//    freopen("5.in","r",stdin);
    memset(f,127/3,sizeof(f));f[0]=0;
    n=read();
    for(int i=1;i<=n;i++) w[i]=read(),deep[i]=read();
    for(int i=n;i>=1;i--) deep[i]+=deep[i+1];
    n++;
    for(int i=1;i<=n;i++) s[i]=w[i]*deep[i];
    for(int i=1;i<=n;i++) p[i]=p[i-1]+s[i];
    for(int i=1;i<=n;i++) k[i]=k[i-1]+w[i];
    for(int i=1;i<=n;i++){
        for(int j=0;j<i;j++){
            f[i]=min(f[i],-deep[j]*k[j]+deep[i]*k[j]);
        }
    }
    int maxn=INT_MAX;
    for(int i=1;i<=n;i++) maxn=min(maxn,f[i]-deep[i]*k[i]+deep[n]*k[i]);cout<<maxn-deep[n]*k[n]+p[n];
}

2

然后再斜率优化一下，因为我的斜率是负数，所以维护一个上凸壳即可

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<climits>
#define int long long
using namespace std;
inline int read(){
    int f=1,ans=0;char c=getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){ans=ans*10+c-‘0‘;c=getchar();}
    return f*ans;
}
const int N=200011;
int w[N],n,f[N],deep[N],sum,s[N],p[N],k[N],l,r,que[N],X[N],Y[N],minn=LLONG_MAX;
signed main(){
//    freopen("5.in","r",stdin);
    memset(f,127/3,sizeof(f));f[0]=0;
    n=read();
    for(int i=1;i<=n;i++) w[i]=read(),deep[i]=read();
    for(int i=n;i>=1;i--) deep[i]+=deep[i+1];
    n++;
    for(int i=1;i<=n;i++) s[i]=w[i]*deep[i];
    for(int i=1;i<=n;i++) p[i]=p[i-1]+s[i];
    for(int i=1;i<=n;i++) k[i]=k[i-1]+w[i];
    l=r=1,que[1]=0;Y[0]=deep[0]*k[0],X[0]=k[0];
    for(int i=1;i<=n;i++){
        while(l<r&&Y[que[l+1]]-Y[que[l]]>=deep[i]*(X[que[l+1]]-X[que[l]])) l++;
        f[i]=deep[i]*k[que[l]]-deep[que[l]]*k[que[l]];
        X[i]=k[i],Y[i]=deep[i]*k[i];
        while(l<r&&(Y[que[r]]-Y[que[r-1]])*(X[i]-X[que[r]])<=(X[que[r]]-X[que[r-1]])*(Y[i]-Y[que[r]])) r--;
        que[++r]=i;
    }
    for(int i=1;i<=n;i++) minn=min(minn,f[i]-deep[i]*k[i]+deep[n]*k[i]);printf("%lld\n",minn-deep[n]*k[n]+p[n]);
    return 0;
}

原文地址：https://www.cnblogs.com/si-rui-yang/p/10119840.html