POJ 2318 TOYS（计算几何）（二分）
大意：给你一个箱子，有n个挡板分隔成n+1部分，给你m个玩具的坐标，问每一部分有几个玩具。
思路：举对每个玩具，二分线段下标，判断玩具在线段左边还是右边，枚举后统计。
1 #include <map> 2 #include <stack> 3 #include <queue> 4 #include <math.h> 5 #include <stdio.h> 6 #include <stdlib.h> 7 #include <string.h> 8 #include <iostream> 9 #include <limits.h> 10 #include <algorithm> 11 #define LL long long 12 #define max(a,b) ((a)>(b)?(a):(b)) 13 #define min(a,b) ((a)<(b)?(a):(b)) 14 #define max3(a, b, c) (a>b?max(a, c):max(b, c)) 15 #define min3(a, b, c) (a<b?min(a, c):min(b, c)) 16 #define max4(a, b, c, d) max(max(a, b), max(c, d)) 17 #define min4(a, b, c, d) min(min(a, b), min(c, d)) 18 #define eps 1e-9 19 #define INF 1 << 30 20 using namespace std; 21 22 int n, m; 23 int cnt[5010]; 24 25 struct Point 26 { 27 int x, y; 28 } P; 29 30 struct node 31 { 32 Point a, b; 33 } Line[5010]; 34 35 int Location(Point a, Point b, Point c) 36 { 37 return (a.x-c.x)*(b.y-c.y) - (b.x-c.x)*(a.y-c.y); 38 } 39 40 void Binary_Search(Point a, int n) 41 { 42 int l, r, mid; 43 l = 0; 44 r = n-1; 45 while(l < r) 46 { 47 mid = (l+r)>>1; 48 if(Location(a, Line[mid].a, Line[mid].b) > 0) 49 { 50 l = mid+1; 51 } 52 else 53 { 54 r = mid; 55 } 56 } 57 if(Location(a, Line[l].a, Line[l].b) < 0) 58 { 59 cnt[l]++; 60 } 61 else 62 { 63 cnt[l+1]++; 64 } 65 } 66 67 void Solve() 68 { 69 int x1, x2, y1, y2, t1, t2; 70 while(~scanf("%d", &n) && n) 71 { 72 memset(cnt, 0, sizeof(cnt)); 73 scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2); 74 for(int i = 0; i < n; ++i) 75 { 76 scanf("%d%d", &t1, &t2); 77 Line[i].a.x = t1; 78 Line[i].a.y = y1; 79 Line[i].b.x = t2; 80 Line[i].b.y = y2; 81 } 82 for(int i = 0; i < m; ++i) 83 { 84 scanf("%d%d", &P.x, &P.y); 85 Binary_Search(P, n); 86 } 87 for(int i = 0; i <= n; ++i) 88 { 89 printf("%d: %d\n", i, cnt[i]); 90 } 91 92 printf("\n"); 93 } 94 } 95 96 int main(void) 97 { 98 freopen("data.in", "r", stdin); 99 //freopen("data.out", "w", stdout); 100 Solve(); 101 102 return 0; 103 }
