【bzoj3165】[Heoi2013]Segment

3165: [Heoi2013]Segment

Time Limit: 40 Sec  Memory Limit: 256 MB
Submit: 718  Solved: 296
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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。  

Input

 
第一行一个整数n，表示共n 个操作。 
接下来n行，每行第一个数为0或1。 
 
若该数为 0，则后面跟着一个正整数 k，表示询问与直线  
x = ((k +lastans–1)%39989+1)相交的线段中交点（包括在端点相交的情形）最靠上的线段的编号，其中%表示取余。若某条线段为直线的一部分，则视作直线与线段交于该线段y坐标最大处。若有多条线段符合要求，输出编号最小的线段的编号。 
若该数为 1，则后面跟着四个正整数 x0, y0, x 1, y 1，表示插入一条两个端点为 
((x0+lastans-1)%39989+1,(y0+lastans-1)%10^9+1)和((x
1+lastans-1)%39989+1,(y1+lastans-1)%10^9+1) 的线段。 
其中lastans为上一次询问的答案。初始时lastans=0。 
 
 

Output

对于每个 0操作，输出一行，包含一个正整数，表示交点最靠上的线段的编号。若不存在与直线相交的线段，答案为0。 

Sample Input

6
1 8 5 10 8
1 6 7 2 6
0 2
0 9
1 4 7 6 7
0 5

Sample Output

2
0 3

HINT

对于100%的数据，1 ≤ n ≤ 10^5 , 1 ≤  k, x0, x1 ≤ 39989, 1 ≤ y0 ≤ y1 ≤ 10^9。

Source

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超哥线段树的区间修改
其实和普通的差不多，只是要在当前区间完全包含于修改区间时才更新
代码：

#include<cstdio>#include<cmath>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;typedef double DL;const int INF = 2147483647;const int maxn = 4 * 40000;const DL eps = 1e-9;int n,tot,ans,len;DL k[maxn],b[maxn];int id[maxn];inline LL getint(){LL ret = 0,f = 1;char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9')ret = ret * 10 + c - '0',c = getchar();return ret * f;}inline bool equal(DL a,DL b){return fabs(a - b) <= eps;}inline void pushdown(int o,int l,int r,int x){DL kx = k[x],bx = b[x],ko = k[id[o]],bo = b[id[o]];int mid = l + r >> 1;DL xx = kx * mid + bx,xo = ko * mid + bo;if (l == r){if (xx > xo) id[o] = x;return;}int lc = o << 1,rc = o << 1 | 1;if (xx < xo || equal(xx,xo)){if (kx < ko) pushdown(lc,l,mid,x);else pushdown(rc,mid + 1,r,x);}else{if (kx < ko) pushdown(rc,mid + 1,r,id[o]);else pushdown(lc,l,mid,id[o]);id[o] = x;}}inline void insert(int o,int l,int r,int al,int ar,int x){if (al <= l && r <= ar){pushdown(o,l,r,x); return;}int mid = l + r >> 1,lc = o << 1,rc = o << 1 | 1;if (al <= mid) insert(lc,l,mid,al,ar,x);if (mid < ar) insert(rc,mid + 1,r,al,ar,x);}inline int qmax(int x,int y,int pos){if (equal(k[x] * pos + b[x],k[y] * pos + b[y])){if (x < y) return x;else return y;}if (k[x] * pos + b[x] < k[y] * pos + b[y]) return y;if (k[x] * pos + b[x] > k[y] * pos + b[y]) return x;}inline int query(int o,int l,int r,int pos){int ret = id[o];if (l == r) return ret;int mid = l + r >> 1,lc = o << 1,rc = o << 1 | 1;if (pos <= mid) ret = qmax(ret,query(lc,l,mid,pos),pos);else ret = qmax(ret,query(rc,mid + 1,r,pos),pos);return ret;}int main(){len = 40000; n = getint(); for (int i = 1; i <= n; i++){int d = getint(); if (!d){//int x = (getint() + ans - 1) % 39989 + 1;int x = getint();printf("%d\n",ans = query(1,1,len,x));}else{//int x0 = (getint() + ans - 1) % 39989 + 1;//int y0 = (getint() + ans - 1) % (int)(1e9) + 1;//int x1 = (getint() + ans - 1) % 39989 + 1;//int y1 = (getint() + ans - 1) % (int)(1e9) + 1;int x0 = getint(),y0 = getint(),x1 = getint(),y1 = getint();if (x0 > x1) swap(x0,x1) , swap(y0,y1);if (x0 == x1){if (y0 > y1) swap(y1,y0); k[++tot] = 0; b[tot] = y1;}else k[++tot] = (y1 - y0) * 1.0 / (x1 - x0) , b[tot] = y0 - k[tot] * x0;insert(1,1,len,x0,x1,tot);}}return 0;}