HDU 5862 Counting Intersections

题意: 问与坐标轴平行的这些线段有多少个交点。

思路： 线扫描的思想，这种做法也是很好想的。

由于坐标的范围较大，可将其中一个坐标离散化（我是离散化X坐标）。

由于题目中有声明：The input data guarantee that no two segments share the same endpoint, no covered segments, and no segments with length 0.

所有没什么坑点。

具体的： 我是将x坐标进行离散化。对于每条边，如果与y轴平行，我分别记录该点的上点和下点，上点标记为-1，下点标记为1，如果是与x平行的边，就直接记录改变，并标记为0；然后对这些点按照y坐标进行排序。 接着就该是加点（边）的操作了。当该点标记为0，就算出(x1,x2)s的和.如果不是0，则在s的x1点加上它的标记。

代码如下：

#include<stdio.h>#include<string.h>#include<math.h>#include<queue>#include<vector>#include<iostream>#include<complex>#include<string>#include<set>#include<map>#include<algorithm>//#include<bits/stdc++.h>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000")#define nn 100100#define mm 2000200#define ll long long#define ULL unsiged long long#define pb push_back#define mod 1000000007#define inf 0xfffffffffff#define eps 0.00000001struct node{int x1, y, x2;int idx;}ax[mm];int posx[mm], s[mm];bool cmp(node a, node b){if(a.y!=b.y) return a.y < b.y;else return a.idx > b.idx;}int lowbit(int x){return x&(-x);}void add(int x,int val,int n){while (x <= n){s[x]+=val;x += lowbit(x);}}int sum(int x){int ans = 0;while (x){ans += s[x];x -= lowbit(x);}return ans;}int main(){int t;//t = io.xint();scanf("%d", &t);while (t--){int n;//n = io.xint();scanf("%d", &n);int nx = 0, ny = 0;for (int i = 0; i < n; i++){int x1, y1, x2, y2;scanf("%d%d%d%d", &x1, &y1, &x2, &y2);//x1 = io.xint(); y1 = io.xint();//x2 = io.xint(); y2 = io.xint();if (x1 > x2) swap(x1, x2);if (y1 > y2) swap(y1, y2);posx[ny++] = x1;posx[ny++] = x2;if (x1 == x2){ax[nx].x1 = x1; ax[nx].y = y1;ax[nx++].idx = 1;ax[nx].x1 = x1; ax[nx].y = y2;ax[nx++].idx = -1;}else{ax[nx].x1 = x1; ax[nx].x2 = x2; ax[nx].y = y1;ax[nx++].idx = 0;}}sort(posx, posx+ny);int lenx = unique(posx, posx + ny) - posx;sort(ax, ax + nx, cmp);memset(s, 0, sizeof(s));ll ans = 0;for (int i = 0; i < nx; i++){if (ax[i].idx == 0){int l = lower_bound(posx, posx + lenx, ax[i].x1) - posx + 1;int r = lower_bound(posx, posx + lenx, ax[i].x2) - posx + 1;ans +=(ll) sum(r) - sum(l - 1);}else{int l = lower_bound(posx, posx + lenx, ax[i].x1) - posx + 1;add(l, ax[i].idx, lenx + 1);}}printf("%lld\n", ans);}return 0;}

记住：要用ll 答案是会超int的。