2018.10.29 bzoj3718: [PA2014]Parking（树状数组）

传送门

显然只用判断两个会相交的车会不会卡住就行了。

直接树状数组维护后缀最大值就行了。

代码：

#include<bits/stdc++.h>
using namespace std;
const int N=5e4+5;
struct Matrix{int x1,x2,w,id;}a1[N],a2[N];
int n,T,W,pos[N],bit[N];
inline int lowbit(int x){return x&-x;}
inline void update(int x,int v){for(int i=x;i;i-=lowbit(i))bit[i]=max(bit[i],v);}
inline int query(int x){int ret=0;for(int i=x;i<=n;i+=lowbit(i))ret=max(ret,bit[i]);return ret;}
inline bool cmp(const Matrix&a,const Matrix&b){return a.x1==b.x1?a.x2<b.x2:a.x1<b.x1;}
const int RLEN=1<<18;
inline char nc() {
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob) ? -1 : *ib++;
}
inline int read() {
    char ch=nc(); int i=0,f=1;
    while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
    while(isdigit(ch)) {i=(i<<1)+(i<<3)+(ch^48); ch=nc();}
    return i*f;
}
int main(){
    T=read();
    while(T--){
        n=read(),W=read();
        for(int i=1,y1,y2;i<=n;++i){
            a1[i].x1=read(),y1=read(),a1[i].x2=read(),y2=read(),a1[i].id=i;
            if(a1[i].x1>a1[i].x2)swap(a1[i].x1,a1[i].x2);
            a1[i].w=abs(y1-y2);
        }
        for(int i=1,y1,y2;i<=n;++i){
            a2[i].x1=read(),y1=read(),a2[i].x2=read(),y2=read(),a2[i].id=i;
            if(a2[i].x1>a2[i].x2)swap(a2[i].x1,a2[i].x2);
            a2[i].w=abs(y1-y2);
        }
        bool f=1;
        sort(a1+1,a1+n+1,cmp),sort(a2+1,a2+n+1,cmp);
        for(int i=1;i<=n;++i)pos[a1[i].id]=i;
        for(int i=1;i<=n;++i){
            if(query(pos[a2[i].id])+a2[i].w>W){f=false;break;}
            update(pos[a2[i].id],a2[i].w);
        }
        fill(bit+1,bit+n+1,0),puts(f?"TAK":"NIE");
    }
    return 0;
}