CodeForces - 12D Ball 线段树

CodeForces - 12D Ball 线段树

图片说明
题意:emmmm,就是题目描述的女性很奇葩,她们去参加国王举办的舞会,而每个女士有B、I、R三种属性,如果她们得知舞会上存在一个比自身所有属性都要高,那么她就会选择跳楼。。。问有多少个要跳楼的女士。

题解:其实很简单,即按照B进行从大到小排序,然后线段树对比I大的区间进行查询最大值,如果最大值大于自身的R那么这位正在查询的女士选择跳楼。最后把这么女士的信息插入到线段树的I点给予R值,即查完再更新。然后查到最后,其中判断成功累积的值就是结果了。

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#include<iostream>
#include<cstring>
#include<sstream>
#include<string>
#include<cstdio>
#include<cctype>
#include<vector>
#include<queue>
#include<cmath>
#include<stack>
#include<list>
#include<set>
#include<map>
#include<algorithm>
#define fi first
#define se second
#define MP make_pair
#define P pair<int,int>
#define PLL pair<ll,ll>
#define Sca(x) scanf("%d",&x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Scl2(x,y) scanf("%lld%lld",&x,&y)
#define Scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define Pri(x) printf("%d\n",x)
#define Prl(x) printf("%lld\n",x)
#define For(i,x,y) for(int i=x;i<=y;i++)
#define _For(i,x,y) for(int i=x;i>=y;i--)
#define FAST_IO std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define ll long long
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;
const double Pi = acos(-1.0);
using namespace std;
template <class T>void tomax(T&a,T b){ a=max(a,b); }
template <class T>void tomin(T&a,T b){ a=min(a,b); }
const int N=5e5+5;
struct Segt{
#define lc (p<<1)
#define rc (p<<1|1)
#define MID (tree[p].l+tree[p].r)>>1
struct Tree{
int l,r;
int val;
}tree[N<<2|1];
void pushdown(int p){ tree[p].val=max(tree[lc].val,tree[rc].val); }
void build(int l,int r,int p){
tree[p]=Tree{l,r,0};
if(l==r) return ;
int mid=MID;
if(l<=mid) build(l,mid,lc);
if(r>mid) build(mid+1,r,rc);
}
void updata(int pos,int p,int val){
if(tree[p].l==pos && tree[p].r==pos){
tomax(tree[p].val,val);
return ;
}
int mid=MID;
if(pos<=mid) updata(pos,lc,val);
if(pos>mid) updata(pos,rc,val);
pushdown(p);
}
int getmax(int l,int r,int p){
if(l>r) return 0;
if(tree[p].l>=l&&tree[p].r<=r) return tree[p].val;
int maxx=0;
int mid=MID;
if(l<=mid) tomax(maxx,getmax(l,r,lc));
if(r>mid) tomax(maxx,getmax(l,r,rc));
return maxx;
}
}t;
struct Data{
int a,b,c;
}num[N];
int tmp[N];

bool cmp(Data one, Data tow){ return one.a > tow.a; }

int discrete(int n){
sort(tmp+1,tmp+1+n);
return unique(tmp+1,tmp+1+n) - tmp;
}

int main(){
int n; Sca(n);
For(i,1,n) Sca(num[i].a);
For(i,1,n) { Sca(num[i].b); tmp[i]=num[i].b; }
For(i,1,n) Sca(num[i].c);
sort(num+1,num+1+n,cmp);
int lenb = discrete(n) - 1;
For(i,1,n) num[i].b=lower_bound(tmp+1,tmp+1+lenb,num[i].b) - tmp;
int ans=0;
t.build(1,lenb,1);
int minn=num[1].a; int l=1;
// For(i,1,n){
// cout<<num[i].a<<' '<<num[i].b<<' '<<num[i].c<<endl;
// }
For(i,2,n){
if(num[i].a<minn){ minn=num[i].a; For(j,l,i-1) t.updata(num[j].b,1,num[j].c); l=i; }
if(t.getmax(num[i].b+1,lenb,1) > num[i].c) ans++;
}
Pri(ans);
}
//9
//7
//7
//7
//5
//2
//1
//1
//1
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