2016暑假多校联合---Counting Intersections
Problem Description
Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection.
The input data guarantee that no two segments share the same endpoint, no covered segments, and no segments with length 0.
The input data guarantee that no two segments share the same endpoint, no covered segments, and no segments with length 0.
Input
The first line contains an integer T, indicates the number of test case.
The first line of each test case contains a number n(1<=n<=100000), the number of segments. Next n lines, each with for integers, x1, y1, x2, y2, means the two endpoints of a segment. The absolute value of the coordinate is no larger than 1e9.
The first line of each test case contains a number n(1<=n<=100000), the number of segments. Next n lines, each with for integers, x1, y1, x2, y2, means the two endpoints of a segment. The absolute value of the coordinate is no larger than 1e9.
Output
For each test case, output one line, the number of intersection.
Sample Input
2
4
1 0 1 3
2 0 2 3
0 1 3 1
0 2 3 2
4
0 0 2 0
3 0 3 2
3 3 1 3
0 3 0 2
Sample Output
4
0
题意：输入n条线段（x1,y1）-（x2,y2） 这n条线段平行于坐标轴，不存在重叠部分且不存在两条线段共断点，求交叉点数；
思路：输入数据n<=100000,而坐标x，y<1e9,所以首先可以将线段进行离散化，使得所有的坐标点在1~1e5之间，把平行于y轴的线段按x坐标排序，把平行于x轴的线段化作起点与结束点，即用一个结构体struct Node{int x；int y；}node[200005]; 保存平行于x轴的线段的信息，例如一条平行于x轴的线段为（x1，y1）-（x2，y2）且y1==y2 则
node[i].x=x1,node[i].y=y1; node[i+1].x=x2+1,node[i+1].y=0-y1; 这样可以把一段区间上的操作化作点的操作，减小复杂度。
区间上的操作化作点的操作，什么意思呢？ 其实就是在这段区间开始的时候加上这个数，那么在区间中间不必再考虑这条线（这段区间），因为开始已经加上了，最后在区间结束的下一位减去就行。
最后，在1到2*1e5的循环，用一个一维的树状数组，当循环到i 时，先把关于node[].x==i 的 操作处理完，然后计算在x=i 直线处相交的点数，即为求出在平行于y轴且x==i的区间上的点数，用树状数组可以方便快速实现；
代码如下：
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <map> using namespace std; typedef long long LL; LL ax[300005],ay[300005]; LL c[200005]; LL Lowbit(LL t) { return t&(t^(t-1)); } LL Sum(LL x) { LL sum = 0; while(x > 0) { sum += c[x]; x -= Lowbit(x); } return sum; } void add(LL li,LL d) { while(li<200005) { c[li]+=d; li=li+Lowbit(li); } } struct Nodex { LL x1,x2; LL y; }nodex[100005]; struct Nodey { LL y1,y2; LL x; }nodey[100005]; struct Node { LL x; LL y; }node[200005]; bool cmp1(const Nodey s1,const Nodey s2) { return s1.x<s2.x; } bool cmp2(const Node s1,const Node s2) { return s1.x<s2.x; } map<LL,LL>q1; map<LL,LL>q2; int main() { LL T,n; scanf("%I64d",&T); while(T--) { q1.clear(); q2.clear(); scanf("%I64d",&n); LL tot1=0,tot2=0; for(LL i=0;i<n;i++) { LL x1,x2,y1,y2; scanf("%I64d%I64d%I64d%I64d",&x1,&y1,&x2,&y2); if(x1==x2) { nodey[tot2].y1=min(y1,y2); nodey[tot2].y2=max(y1,y2); nodey[tot2++].x=x1; } else { nodex[tot1].x1=min(x1,x2); nodex[tot1].x2=max(x1,x2); nodex[tot1++].y=y1; } } LL num1,num2; for(LL i=0;i<tot1;i++) { ax[i*2]=nodex[i].x1; ax[i*2+1]=nodex[i].x2; } num1=2*tot1; for(LL i=0;i<tot2;i++) { ax[num1++]=nodey[i].x; } sort(ax,ax+num1); LL tot=0,pre=-1; for(LL i=0;i<num1;i++) { if(ax[i]!=pre) { pre=ax[i]; q1[pre]=++tot; } } for(LL i=0;i<tot2;i++) { ay[i*2]=nodey[i].y1; ay[i*2+1]=nodey[i].y2; } num2=2*tot2; for(LL i=0;i<tot1;i++) ay[num2++]=nodex[i].y; sort(ay,ay+num2); tot=0,pre=-1; for(LL i=0;i<num2;i++) { if(ay[i]!=pre) { pre=ay[i]; q2[pre]=++tot; } } for(LL i=0;i<tot1;i++) { nodex[i].x1=q1[nodex[i].x1]; nodex[i].x2=q1[nodex[i].x2]; nodex[i].y=q2[nodex[i].y]; } for(LL i=0;i<tot2;i++) { nodey[i].y1=q2[nodey[i].y1]; nodey[i].y2=q2[nodey[i].y2]; nodey[i].x=q1[nodey[i].x]; } //for(LL i=0;i<tot1;i++) //cout<<"x: "<<nodex[i].x1<<" "<<nodex[i].x2<<" "<<nodex[i].y<<endl; //for(LL i=0;i<tot1;i++) //cout<<"y: "<<nodey[i].y1<<" "<<nodey[i].y2<<" "<<nodey[i].x<<endl; sort(nodey,nodey+tot2,cmp1); tot=0; for(LL i=0;i<tot1;i++) { node[tot].x=nodex[i].x1; node[tot++].y=nodex[i].y; node[tot].x=nodex[i].x2+1; node[tot++].y=0-nodex[i].y; } sort(node,node+tot,cmp2); LL j=0,k=0,sum=0; memset(c,0,sizeof(c)); for(LL i=1;i<=200001;i++) { while(node[j].x==i&&j<tot) { if(node[j].y>0) add(node[j].y,1); else add(0-node[j].y,-1); j++; } while(nodey[k].x==i&&k<tot2) { sum+=Sum(nodey[k].y2)-Sum(nodey[k].y1-1); k++; } if(j>=tot) break; if(k>=tot2) break; } printf("%I64d\n",sum); } return 0; }
