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HDU 3465 Life is a Line (逆序数 + 数状数组)

2017-07-05 16:18 471 查看

Life is a Line

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 217 Accepted Submission(s): 81
 
Problem DescriptionThere is a saying: Life is like a line, some people are your parallel lines, while others are destined to meet you.Maybe have met, maybe just a matter of time, two unparallel lines will always meet in some places, and now a lot of life (i.e. line) are in the same coordinate system, in a given open interval, how many pairs can meet each other? 
InputThere are several test cases in the input.Each test case begin with one integer N (1 ≤ N ≤ 50000), indicating the number of different lines.Then two floating numbers L, R follow (-10000.00 ≤ L < R ≤ 10000.00), indicating the interval (L, R).Then N4000lines follow, each line contains four floating numbers x1, y1, x2, y2 (-10000.00 ≤ x1, y1, x2, y2 ≤ 10000.00), indicating two different points on the line. You can assume no two lines are the same one.The input terminates by end of file marker. 
OutputFor each test case, output one integer, indicating pairs of intersected lines in the open interval, i.e. their intersection point’s x-axis is in (l, r). 
Sample Input
3
0.0 1.0
0.0 0.0 1.0 1.0
0.0 2.0 1.0 2.0
0.0 2.5 2.5 0.0
 
Sample Output
1
问题：求在 （l, r）范围内有多少个交点，看起来好像是个几何题。。。时间上是一个求逆序数的问题。
     只要直线不平行于y轴就一定会 跟 x = l, x = r这两条直线相交。对于任意两条直线x=l,和x=r的交点分别为xl,xr，两直线在(L,R)相交，必然有 xl1>xl2&&xr1<xr2 或者 xl1<xl2&&xr1>xr2，(A.L-B.L)*(A.R-B.R)<0 所以可以用逆序数来求解。如果将与L的交点从小到大排,那么当相应的R的交点出现一个逆序数对就表示有两直线有交点.  先将所有直线根据l递增排序，之后编号1~n，再根据r递减排序，得到一个编号序列。  例如3412，递减，其中r3>r4>r1>r2, 又编号：l1<l2<r3<r4,1234=>所以12 34 所以3和4有交点，1和2有交点。  这符合逆序数的关系：一个数的逆序数是在它之前比他大的数的个数，  当然这里是小的数，原因是为了方便树状数组处理。所以只要根据上述方法排序再求逆序数即可。  1.与y轴平行得线，只要这样的线在(l,r)范围内，则必定跟别的不平行线相交。     所以计算下个数在乘积  2.l和r相同的情况。当l相同时，r递增排序；当r相同时，l递减排序。     就能使在l和r上的交点不计算在内。  （这一段转的别人的）
#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int maxn = 50000;struct Line{double a, b;int num;}line[50050];int tree[50500];int lowbit(int x) {return x & (-x);}void add(int pos) {for (int i = pos; i <= maxn; i += lowbit(i)) {tree[i]++;}}int sum(int pos) {int ans = 0;for (int i = pos; i >= 1; i -= lowbit(i)) {ans += tree[i];}return ans;}int main() {int n;double left, right;double x1, x2, y1, y2;while (cin >> n) {int parallel_y = 0;int unparallel = 0;scanf("%lf%lf", &left, &right);for (int i = 0; i < n; ++i) {scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);if (x1 == x2) {if (x1 > left && x1 < right) {parallel_y++;}} else {double k = (y1 - y2) / (x1 - x2);double b = y1 - k * x1;line[unparallel].a = left * k + b;line[unparallel++].b = right * k + b;}}sort(line, line + unparallel, [](Line l1, Line l2)->bool{return l1.a == l2.a ? l1.b < l2.b : l1.a < l2.a;});for (int i = 0; i < unparallel; ++i) {line[i].num = i + 1;}sort(line, line + unparallel, [](Line l1, Line l2)->bool{return l1.b == l2.b ? l1.a > l2.a : l1.b > l2.b;});memset(tree, 0, sizeof(tree));int ans = 0;for (int i = 0; i < unparallel; ++i) {add(line[i].num);ans += sum(line[i].num - 1);}cout << ans + unparallel * parallel_y << endl;}}

                                            

                                        

                        
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