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zc
2019-12-21 19:47:00
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f_i为前i个任务的费用,t_i为前i个任务时间的总和,w_i为前i个任务费用的总和

f_i=\min_{j=0}^{i-1}(f_j+s\times(w_n-w_j)+t_i\times(w_i-w_j))

设任意0\le j<k<ij转移比从k转移优,那么

f_j+s\times(w_n-w_j)+t_i\times(w_i-w_j) \le f_k+s\times(w_n-w_k)+t_i\times(w_i-w_k) \\ t_i(w_k-w_j)\le f_k-f_j+s(w_j-w_k) \\ t_i\le \frac{f_k-f_j+s(w_j-w_k)}{w_k-w_j}

维护了个下凸壳交了上去,结果听取\color{red}\text{WA}声一片

仔细看了看|T_i|\le 2^8

这也就意味着我们每次截取的直线的斜率不再单调递增,单调队列不能再满足需要

为了保证决策状态的有序性,我们还是选择用一个单调栈维护下凸包。

因为决策集合已经有序,对于任意直线,我们只用在队列中二分出使截距最小的交点

由于出题人十分毒瘤,我们要把斜率大小的判断从相除改成十字相乘

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 300011
#define int ll
int s,n,q[N],tp=0;
ll f[N],t[N],w[N];
bool chk(int i,int j,int I,int J){
    return (f[j]-f[i]+s*(w[i]-w[j]))*(w[J]-w[I])>(f[J]-f[I]+s*(w[I]-w[J]))*(w[j]-w[i]);
}
bool CHK(int i,int j,int v){
    return f[j]-f[i]+s*(w[i]-w[j])>v*(w[j]-w[i]);
}
int solve(int v){
    int l=0,r=tp-1,ans=-1;
    while(l<=r){
        int m=(l+r)>>1;
        if(CHK(q[m],q[m+1],v))ans=m,r=m-1;
        else l=m+1;
    }
    if(~ans)return q[ans];
    else return q[tp];
}
signed main(){
    in(n,s);
    Fur(i,1,n)
        in(t[i],w[i]),
        t[i]+=t[i-1],
        w[i]+=w[i-1];

    clr(f,0x7f);
    f[0]=q[0]=0;
    Fur(i,1,n){
        int j=solve(t[i]);
        f[i]=f[j]+s*(w[n]-w[j])+t[i]*(w[i]-w[j]);
        while(tp&&chk(q[tp-1],q[tp],q[tp],i))--tp;
        q[++tp]=i;
    }
    cout<<f[n]<<endl;
}
LG 5785 [SDOI2012]任务安排
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