Counting Intersections

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题意：给出平行于坐标轴的线段，保证无重叠，端点相同的情况，问有多少交点。
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解法：首先我把数离散化，那么数的大小不超过4*10^5，这样就可以用树状数组。首先把平行于x、y轴的分开存。对于垂直于x轴的按x坐标排序，对于垂直于y轴的按x坐标左端点进行排序。然后扫描垂直x坐标轴的线段，可以很快找到所以垂直y轴的而且左端点在该x坐标左边的所有线段（显然这个扫描是线性的），我把这些线段放到堆中维护（代码中用了heap），以右端点为关键字，那么可以很快剔除所有右端点在x坐标左边的、即不合法没有与这条线段相交的线段（同样是线性的），于是我们得到了所有垂直y轴而且横跨过该x坐标的所有线段，用树状数组维护就好了。
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#include <iostream>#include <stdlib.h>#include <stdio.h>#include <algorithm>#include <map>#include <string.h>using namespace std;const int maxn = 200010;int input[maxn][5], num, tot, tot_x, tot_y, n, t;long long ans, f[4*maxn];struct Node{    int num, x, y;    bool operator<(Node &other) {        return num < other.num;    }}to_sort[4*maxn];struct dy {    int y, x1, x2;    bool operator<(const dy &other) const {        if (x2 == other.x2)        {            if (x1 == other.x1)            {                return y < other.y;            }            return x1 < other.x1;        }        return x2 < other.x2;    }}y_a[maxn];bool cmp(dy t1, dy t2) {    return t1.x1 < t2.x1;}struct dx {    int x, y1, y2;    bool operator<(const dx &other)const {        return x < other.x;    }}x_a[maxn];map<dy, int> heap;void insert(int x) {    while (1) {        f[x] = f[x] + 1ll;        x = x+(x & (-x));        if (x > num) break;    }}void cancel(int x) {    while (1) {        f[x] = f[x] - 1ll;        x = x+(x & (-x));        if (x > num) break;    }}long long find(int x) {    long long sum = 0;    while (1) {        if (x == 0) break;        sum = sum+f[x];        x = x-(x & (-x));    }    return sum;}int main() {    int tt;    scanf("%d", &tt);    while (tt--) {        tot = 0;        scanf("%d", &n);        for (int i = 1; i <= n; i++)            for (int j = 1; j <= 4; j++) {                tot++;                scanf("%d", &to_sort[tot].num);                to_sort[tot].x = i;                to_sort[tot].y = j;            }        sort(to_sort+1, to_sort+tot+1);        num = 1;        input[to_sort[1].x][to_sort[1].y] = num;        for (int i = 2; i <= n*4; i++) {            if (to_sort[i].num != to_sort[i-1].num) num++;            input[to_sort[i].x][to_sort[i].y] = num;        }        tot_x = 0;        tot_y = 0;        for (int i = 1; i <= n; i++) {            if (input[i][1] == input[i][3]) {                tot_x++;                x_a[tot_x].x = input[i][1];                x_a[tot_x].y1 = min(input[i][2], input[i][4]);                x_a[tot_x].y2 = max(input[i][2], input[i][4]);            } else if (input[i][2] == input[i][4]) {                tot_y++;                y_a[tot_y].y = input[i][2];                y_a[tot_y].x1 = min(input[i][1], input[i][3]);                y_a[tot_y].x2 = max(input[i][1], input[i][3]);            }        }        sort(x_a+1, x_a+tot_x+1);        sort(y_a+1, y_a+tot_y+1, cmp);        if (tot_x+tot_y != n) while (1) {}        heap.clear();        memset(f, 0, sizeof(f));        t = 1;        ans = 0;        for (int i = 1; i <= tot_x; i++) {            while (t <= tot_y && y_a[t].x1 <= x_a[i].x) {                heap[y_a[t]] = 1;                insert(y_a[t].y);                t++;            }            while (1) {                if (heap.size() == 0) break;                map<dy, int>::iterator k = heap.begin();                if (k->first.x2 < x_a[i].x) {                    cancel(k->first.y);                    heap.erase(k);                } else break;            }            ans = ans+find(x_a[i].y2)-find(x_a[i].y1-1);        }        printf("%lld\n", ans);    }}