2017.10.4 QBXT 模拟赛

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;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;top--)
        {
            R[i]+=a[num[stack[top]]];
            ans=max(ans,R[i]);
        }
        stack[++top]=i;
    }
    cout<<ans<<endl;
    return 0;
}

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;
}

T3

  解析几何

#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;
    }
}