题目:

Description

要求在平面直角坐标系下维护两个操作：

1. 在平面上加入一条线段。记第i条被插入的线段的标号为i。
2. 给定一个数k,询问与直线 x = k相交的线段中，交点最靠上的线段的编号。

题解:

同[SDOI 2016]游戏
还要简单一些,就不写题解了.

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(ll &x){x=0;char ch;bool flag = false;while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const ll maxn = 100010;
const ll maxp = 40010;
const double eps = 1e-9;
inline ll dcmp(double x){return (x > eps) - (x < -eps);
}
struct Node{double k,b;ll id;bool vis;Node(){id = 0;k = b = 0;}
}T[maxn<<2];
ll L,R,idx;double K,B;
inline void solve(ll rt,ll l,ll r){if(T[rt].vis == false){T[rt].k = K;T[rt].b = B;T[rt].id = idx;T[rt].vis = true;return ;}double y0 = T[rt].k*l + T[rt].b;double y1 = K*l + B;double y2 = T[rt].k*r + T[rt].b;double y3 = K*r + B;if(dcmp(y1-y0) > 0 && dcmp(y3 - y2) > 0){if(dcmp(K-T[rt].k) == 0 && dcmp(B - T[rt].b) == 0) return;T[rt].k = K;T[rt].b = B;T[rt].id = idx;return ;}if(dcmp(y1-y0) <= 0 && dcmp(y3 - y2) <= 0) return ;ll mid = l+r >> 1;solve(rt<<1,l,mid);solve(rt<<1|1,mid+1,r);
}
inline void insert(ll rt,ll l,ll r){if(L <= l && r <= R){solve(rt,l,r);return ;}ll mid = l+r >> 1;if(L <= mid) insert(rt<<1,l,mid);if(R >  mid) insert(rt<<1|1,mid+1,r);
}
inline void insert(ll x0,ll x1,ll y0,ll y1,ll num){L = x0;R = x1;K = (double)(y1-y0)/(double)(x1-x0);B = y0 - x0*K;idx = num;insert(1,1,maxp);
}
typedef pair<ll,ll> pa;
double ans;ll ans_id;
inline void query(ll rt,ll l,ll r,ll pos){double x = T[rt].k*pos + T[rt].b;if(dcmp(x-ans) == 1 || (dcmp(x-ans) == 0 && T[rt].id < ans_id)) ans = x,ans_id = T[rt].id;if(l == r) return ;ll mid = l+r >> 1;if(pos <= mid) query(rt<<1,l,mid,pos);else query(rt<<1|1,mid+1,r,pos);
}
ll a[maxn],id[maxn];
int main(){ll n;read(n);ll lastans = 0;ll op,x0,y0,x1,y1,x;ll num = 0;while(n--){read(op);if(op == 0){read(x);x = ((x +lastans-1)%39989+1);ans = ans_id = 0;query(1,1,maxp,x);if(ans < a[x] || (ans == a[x] && id[x] < ans_id)) ans_id = id[x];printf("%d\n",lastans = ans_id);}else if(op == 1){++ num;read(x0);read(y0);read(x1);read(y1);x0 = (x0+lastans-1)%39989+1;y0 = (y0+lastans-1)%1000000000+1;x1 = (x1+lastans-1)%39989+1;y1 = (y1+lastans-1)%1000000000+1;if(x0 > x1) swap(x0,x1),swap(y0,y1);if(x0 == x1 ){if(max(y0,y1) > a[x0]) a[x0] = max(y0,y1),id[x0] = num;}else insert(x0,x1,y0,y1,num);}}   getchar();getchar();return 0;
}