2017.10.4 QBXT 模拟赛
分类:
IT文章
•
2022-03-02 23:53:12
题目链接
T1
维护一个单调栈
#include <iostream>
#include <cstdio>
#define N 500000
#define rep(a,b,c) for(int a=b;a<=c;++a)
#define Rep(a,b,c) for(int a=b;a>=c;--a)
using namespace std;
typedef long long LL;
int n,top,stack[N],num[N];
LL ans,h[N],a[N],R[N];
int main()
{
cin>>n;
rep(i,1,n)
{
cin>>h[i]>>a[i],num[i]=i;
for(;h[stack[top]]<h[i]&⊤top--)
{
R[i]+=a[num[stack[top]]];
ans=max(ans,R[i]);
}
stack[++top]=i;
}
top=0;
Rep(i,n,1)
{
for(;h[stack[top]]<h[i]&⊤top--)
{
R[i]+=a[num[stack[top]]];
ans=max(ans,R[i]);
}
stack[++top]=i;
}
cout<<ans<<endl;
return 0;
}
View Code
T2
二分
二分矩形的边长,然后我们再试着枚举左边界,通过左边界我们就可以知道右边界,同样通过右边界我们也可以知道左边界,然后我们再在通过左右边界来判断里面的糖果数是否足够,当然我们的上下也是有边界的,我们在当前左右所加的这个矩形里,如果糖果数不够c那么直接返回false,反之,我们在来判断一下在这个区间里每个糖果的y坐标排序,然后将这个糖果的y与他的前c个糖果的y比较,如果小于等于mid的话,说明在这样的长度里我们可以找到c个糖果。
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
struct node {int x,y;} a[1005];
int C,n,L,R,mid,b[1005],o,i;
int cmp(node i,node j) {return i.x<j.x;}
int CMP(int i,int j) {return i<j;}
bool WORK(int l,int r)
{
if (r-l+1<C) return false; o=0;
for (int i=l; i<=r; i++) b[++o]=a[i].y;
sort(b+1,b+o+1,CMP);
for (int i=C; i<=o; i++)
if (b[i]-b[i-C+1]<=mid) return true;
return false;
}
bool OK(int x)
{
int l=1;
for (int i=1; i<=n; i++)
{
if (a[i].x-a[l].x>x)
{
if (WORK(l,i-1)) return true;
while (a[i].x-a[l].x>x) l++;
}
}
if (WORK(l,n)) return true;
return false;
}
int main()
{
freopen("square.in","r",stdin);
freopen("square.out","w",stdout);
scanf("%d%d",&C,&n);
for (i=1; i<=n; i++)
scanf("%d%d",&a[i].x,&a[i].y);
sort(a+1,a+n+1,cmp);
L=0; R=10000; mid=(L+R)/2;
while (L<=R)
{
if (OK(mid)) {R=mid-1; mid=(L+R)/2;} else
{
L=mid+1;
mid=(L+R)/2;
}
}
cout<<L+1;
return 0;
}
View Code
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const long double INF=(long double)1000000000*10;
long double L,R,mid,ans,hh[100005];
int r,rr,i,n,MAX,X,Y,cnt,vv[100005],vv2[100005];
struct node2 {int t; long double l;} s[200005],S[200005];
struct node {int t,v;} t[100005];
int cmp(node i,node j) {return i.v<j.v || i.v==j.v && i.t>j.t;}
struct Node {long double x;int y,z;} p[200005];
int CMP(Node i,Node j) {return i.x<j.x;}
long double work(int x,long double y) {return (long double)t[x].v*y-hh[x];}
int main()
{
freopen("chase.in","r",stdin);
freopen("chase.out","w",stdout);
while (1)
{
scanf("%d",&n);
// if (n==0) return 0;
MAX=0;
for (i=1; i<=n; i++)
{
scanf("%d%d",&t[i].t,&t[i].v);
MAX=max(MAX,t[i].t);
}
sort(t+1,t+n+1,cmp); int MIN=t[n].t;
for (i=n-1; i>=2; i--)
{
if (t[i].t>MIN) vv[i]=1; else
MIN=t[i].t,vv[i]=0;
}
for (i=1; i<=n; i++) hh[i]=(long double)t[i].t*t[i].v;
r=1; s[1].l=MAX; s[1].t=1; s[2].l=INF; vv[n]=0;
for (i=2; i<=n; i++)
if (!vv[i])
{
while (r && work(i,s[r].l)>=work(s[r].t,s[r].l)) r--;
if (!r) {r=1; s[1].l=MAX; s[1].t=i; continue;}
L=s[r].l; R=s[r+1].l; mid=(L+R)/2.0;
for (int I=1; I<=80; I++)
{
if (work(i,mid)>=work(s[r].t,mid)) {R=mid; mid=(L+R)/2.0;} else {L=mid; mid=(L+R)/2.0;}
}
s[++r].l=mid; s[r].t=i; s[r+1].l=INF;
}
rr=1; S[1].l=MAX; S[2].l=INF; S[1].t=n;
MIN=t[1].t;
for (i=2; i<n; i++)
if (t[i].t<MIN) vv2[i]=1; else
MIN=t[i].t,vv2[i]=0;
for (i=n-1; i>=1; i--)
if (!vv2[i])
{
while (rr && work(i,S[rr].l)<=work(S[rr].t,S[rr].l)) rr--;
if (!rr) {rr=1; S[1].l=MAX; S[1].t=i; continue;}
L=S[rr].l; R=S[rr+1].l; mid=(L+R)/2.0;
for (int I=1; I<=80; I++)
{
if (work(i,mid)<=work(S[rr].t,mid)) {R=mid; mid=(L+R)/2.0;} else {L=mid; mid=(L+R)/2.0;}
}
S[++rr].l=mid; S[rr].t=i; S[rr+1].l=INF;
}
cnt=0;
for (i=1; i<=r; i++) {p[++cnt].x=s[i].l; p[cnt].y=1; p[cnt].z=s[i].t;}
for (i=1; i<=rr; i++) {p[++cnt].x=S[i].l; p[cnt].y=0; p[cnt].z=S[i].t;}
sort(p+1,p+cnt+1,CMP); X=Y=0; ans=INF;
for (i=1; i<=cnt; i++)
{
if (p[i].y==1) X=p[i].z; else Y=p[i].z;
// printf("%.5f
",(double)p[i].x);
if (X && Y) ans=min(ans,work(X,p[i].x)-work(Y,p[i].x));
}
printf("%.2f
",fabs((double)ans));
return 0;
}
}