POJ 2318：
题目大意：给定一个盒子的左上角和右下角坐标，然后给n条线，可以将盒子分成n+1个部分，再给m个点，问每个区域内有多少各点
这个题用到关键的一步就是向量的叉积，假设一个点m在 由abcd围成的四边形区域内，那么向量ab, bc, cd, da和点的关系就是，点都在他们的同一侧，我是按照逆时针来算的，所以只需要判断叉积是否小于0就行了。还有一个问题就是这个题要求时间是2s，所以直接找，不用二分也能过，不过超过1s了，最好还是二分来做，二分时间170ms，二分的时候要把最左边的一条边和最右边的一条边都要加到数组中去
代码一(直接找区域)
#include<iostream> #include <cstdio> #include <string.h> using namespace std; const int N = 15500; struct point{ int x, y; }; int n, m, ans[5500]; point upper_left, lower_left, upper_right, lower_right, p[N]; double direction(point a, point b, point c)//判断方向 { point t1, t2; t1.x = c.x - a.x; t1.y = c.y - a.y; t2.x = b.x - a.x; t2.y = b.y - a.y; return (t1.x * t2.y * 1.0 - t1.y * t2.x * 1.0); } int main() { while (~scanf("%d", &n) && n) { memset(ans, 0, sizeof(ans)); int a, b; scanf("%d %d %d %d %d", &m, &upper_left.x, &upper_left.y, &lower_right.x, &lower_right.y); lower_left.x = upper_left.x; lower_left.y = lower_right.y; upper_right.x = lower_right.x; upper_right.y = upper_left.y; for (int i = 0; i < 2 * n; i++) { scanf("%d %d", &a, &b); p[i].x = a; p[i].y = upper_left.y; p[++i].x = b; p[i].y = lower_right.y; } point tmp; double d1, d2, d3, d4; for (int i = 0; i < m; i++) { scanf("%d %d", &tmp.x, &tmp.y); //判断是否在第一个区域内 d1 = direction(upper_left, lower_left, tmp); d2 = direction(lower_left, p[1], tmp); d3 = direction(p[1], p[0], tmp); d4 = direction(p[0], upper_left, tmp); if (d1 <= 0 && d2 <= 0 && d3 <= 0 && d4 <= 0) { ans[0]++; continue; } bool flag = false; for (int j = 0; j < 2 * n - 2; j += 2) { d1 = direction(p[j], p[j + 1], tmp); d2 = direction(p[j + 1], p[j + 3], tmp); d3 = direction(p[j + 3], p[j + 2], tmp); d4 = direction(p[j + 2], p[j], tmp); if (d1 <= 0 && d2 <= 0 && d3 <= 0 && d4 <= 0) { ans[j / 2 + 1]++; flag = true; break; } } if (flag) continue; //d1 = direction(p[2 * n - 2], p[2 * n - 1], tmp); //d2 = direction(p[2 * n - 1], lower_right, tmp); //d3 = direction(lower_right, upper_right, tmp); //d4 = direction(upper_right, p[n * 2 - 2], tmp); //if (d1 <= 0 && d2 <= 0 && d3 <= 0 && d4 <= 0) ans[n]++; } for (int i = 0; i <= n; i++) printf("%d: %d ", i, ans[i]); puts(""); } return 0; }
代码二(二分法)
#include<iostream> #include <cstdio> #include <string.h> using namespace std; const int N = 15500; struct point{ int x, y; }; int n, m, ans[5500]; point upper_left, lower_left, upper_right, lower_right, p[N];//各个角 double direction(point a, point b, point c)//判断方向，判断点c在向量ab的哪一侧，如果返回是负值则在逆时针那侧 { point t1, t2; t1.x = c.x - a.x; t1.y = c.y - a.y; t2.x = b.x - a.x; t2.y = b.y - a.y; return (t1.x * t2.y * 1.0 - t1.y * t2.x * 1.0); } void dichotomization(point tmp)//二分法判断在那块区间内 { int left, right, mid; left = 0; right = n + 1; mid = (left + right) >> 1; while (left < right) { if (left + 1 == right) { ans[left]++; break; } double d = direction(p[mid * 2], p[mid * 2 + 1], tmp); if (d < 0) left = mid; else right = mid; mid = (left + right) >> 1; } } int main() { while (~scanf("%d", &n) && n) { memset(ans, 0, sizeof(ans)); int a, b; scanf("%d %d %d %d %d", &m, &upper_left.x, &upper_left.y, &lower_right.x, &lower_right.y); lower_left.x = upper_left.x; lower_left.y = lower_right.y; upper_right.x = lower_right.x; upper_right.y = upper_left.y; p[0] = upper_left;//将边角点加入p数组，为了二分 好计算 p[1] = lower_left; for (int i = 2; i < 2 * n + 2; i++) { scanf("%d %d", &a, &b); p[i].x = a; p[i].y = upper_left.y; p[++i].x = b; p[i].y = lower_right.y; } p[2 * n + 2] = upper_right; p[2 * n + 3] = lower_right; point tmp; for (int i = 0; i < m; i++) { scanf("%d %d", &tmp.x, &tmp.y); dichotomization(tmp); } for (int i = 0; i <= n; i++) printf("%d: %d ", i, ans[i]); puts(""); } return 0; }
POJ 2398：
这道题和上一题类似，包括主要过程也一样，但是有个条件不一样，上一个题它给的n条线时有序的，就是从左往右的，但是这个没有顺序，是随便给的，所以要把它排一下序，最后让求的，假设是一个区域内有t个点，的这样的区域有多少个，按照升序打印出来：
代码如下：
#include<iostream> #include <cstdio> #include <string.h> #include <algorithm> using namespace std; const int N = 5500; struct point{ int x, y; }; struct Edge{ point Start, End; }; Edge edge[N]; int n, m, ans[N], res[N]; point upper_left, lower_left, upper_right, lower_right, p[N];//各个角 bool cmp(Edge e1, Edge e2) { if (e1.Start.x != e2.Start.x) return e1.Start.x < e2.Start.x; return e1.End.x < e2.End.x; } double direction(point a, point b, point c)//判断方向，判断点c在向量ab的哪一侧，如果返回是负值则在逆时针那侧 { point t1, t2; t1.x = c.x - a.x; t1.y = c.y - a.y; t2.x = b.x - a.x; t2.y = b.y - a.y; return (t1.x * t2.y * 1.0 - t1.y * t2.x * 1.0); } void dichotomization_edge(point tmp) { int left, right, mid; left = 0; right = n + 1; mid = (left + right) >> 1; while (left < right) { if (left + 1 == right) { ans[left]++; break; } double d = direction(edge[mid].Start, edge[mid].End, tmp); if (d < 0) left = mid; else right = mid; mid = (left + right) >> 1; } } int main() { while (~scanf("%d", &n) && n) { memset(ans, 0, sizeof(ans)); point t1, t2; scanf("%d %d %d %d %d", &m, &upper_left.x, &upper_left.y, &lower_right.x, &lower_right.y); lower_left.x = upper_left.x; lower_left.y = lower_right.y; upper_right.x = lower_right.x; upper_right.y = upper_left.y; edge[0].Start = upper_left; edge[0].End = lower_left; for (int i = 1; i <= n; i++) { scanf("%d %d", &t1.x, &t2.x); t1.y = upper_left.y; t2.y = lower_right.y; edge[i].Start = t1; edge[i].End = t2; } edge[n + 1].Start = upper_right; edge[n + 1].End = lower_right; point tmp; sort(edge, edge + n + 1, cmp); for (int i = 0; i < m; i++) { scanf("%d %d", &tmp.x, &tmp.y); dichotomization_edge(tmp); } memset(res, 0, sizeof(res)); for (int i = 0; i <= n; i++) if (ans[i] != 0) res[ans[i]]++; printf("Box "); for (int i = 0; i < 5000; i++) if (res[i] != 0) printf("%d: %d ", i, res[i]); } return 0; }