JZOJ5390. 【NOIP2017提高A组模拟9.26】逗气

题解

很显然，我们将一个逗气阵变为求左边的最大值和右边的最大值，这样就能去掉绝对值。
点和点之间的斜率是不变的，那就可以用这个斜率维护一个凸壳。
对于每一个聚集点，就将它加入凸壳。
对于每一个聚逗阵，就二分一定点，使它的斜率<di

code

#include<queue>#include<cstdio>#include<iostream>#include<algorithm>#include <cstring>#include <string.h>#include <cmath>#include <math.h>#define ll long long#define N 200003#define db double#define P putchar#define G getchar#define mo 1000000007using namespace std;char ch;void read(ll &n){    n=0;    ch=G();    while((ch<'0' || ch>'9') && ch!='-')ch=G();    ll w=1;    if(ch=='-')w=-1,ch=G();    while('0'<=ch && ch<='9')n=(n<<3)+(n<<1)+ch-'0',ch=G();    n*=w;}ll max(ll a,ll b){return a>b?a:b;}ll min(ll a,ll b){return a<b?a:b;}void write(ll x){     if(x>9) write(x/10);     P(x%10+'0');}struct node{    ll x,y;    int z,op; }a[N*2];bool cmp1(node a,node b){return a.x<b.x || (a.x==b.x && a.op<b.op);}bool cmp2(node a,node b){return a.x>b.x || (a.x==b.x && a.op<b.op);}int n,m,top,z[N],l,r,mid,x;ll ans[N];db xl(int x,int y){    return (db)(a[x].y-a[y].y)*1.000000/(db)(a[x].x-a[y].x);}int main(){    freopen("gas.in","r",stdin);    freopen("gas.out","w",stdout);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)        read(a[i].x),read(a[i].y),a[i].op=1;    for(int i=1;i<=m;i++)        read(a[i+n].x),read(a[i+n].y),a[i+n].z=i,a[i+n].op=2;    sort(a+1,a+1+n+m,cmp1);    for(int i=1;i<=n+m;i++)    {        if(a[i].op==1)        {            while(top>1 && xl(z[top-1],z[top])<=xl(z[top],i))top--;            z[++top]=i;        }        else        {            if(top==0)continue;            l=1;r=top-1;x=1;            while(l<=r)            {                mid=(l+r)>>1;                if(xl(z[mid],z[mid+1])>=(db)-a[i].y)x=mid+1,l=mid+1;else r=mid-1;            }            ans[a[i].z]=max(ans[a[i].z],a[z[x]].y-(a[i].x-a[z[x]].x)*a[i].y);        }    }    sort(a+1,a+1+n+m,cmp2);    top=0;memset(z,0,sizeof(z));    for(int i=1;i<=n+m;i++)        if(a[i].op==1)        {            while(top>1 && xl(z[top-1],z[top])>=xl(z[top],i))top--;            z[++top]=i;        }        else        {            if(top==0)continue;            l=1;r=top-1;x=1;            while(l<=r)            {                mid=(l+r)>>1;                if(xl(z[mid],z[mid+1])<(db)a[i].y)x=mid+1,l=mid+1;else r=mid-1;            }            ans[a[i].z]=max(ans[a[i].z],a[z[x]].y-(a[z[x]].x-a[i].x)*a[i].y);        }    for(int i=1;i<=m;i++)        write(ans[i]),P('\n');}