HDU 4380 Farmer Greedy calculation ry + bitset

HDU 4380 Farmer Greedy calculation ry + bitset

Enumerative straight lines. If a point in a straight line is left or right, You can compress a number and use bitset to record it.

Then the triangle is three bitsets & bit

#include 
 
  #include 
  
   #include 
   
    #include using namespace std;typedef long long ll;const int N = 101;const int M = 1005;bitset
    
      b1[N*N], b2[N*N];int x[N], y[N], px[M], py[M];bool check(int i, int j, int k) {if(x[i] != x[j]) {double yy = (double)(y[i] - y[j]) / (x[i] - x[j]) * (x[k] - x[i]) + y[i];if(y[k] >= yy) return true;else return false;} else {  if(x[k] >= x[i])return true;else return false;}}void put(bitset
     
       x) {for(int i = 0; i < M; i ++) {if(x[i]) printf("%d ", i);}puts("*");}int main() {int n, m, cas = 0;while(~scanf("%d%d", &n, &m)) {for(int i = 0; i < n; i ++) {scanf("%d%d", &x[i], &y[i]);}for(int i = 0; i < m; i ++) {scanf("%d%d", &px[i], &py[i]);}for(int i = 0; i < n; i ++) {for(int j = i+1; j < n; j ++) {for(int k = 0; k < m; k ++) {//printf("%d %d  ", i, j);if(x[i] != x[j]) {double yy = (double)(y[i] - y[j]) / (x[i] - x[j]) * (px[k] - x[i]) + y[i];if(py[k] == yy) {b1[i*n+j].set(k);b2[i*n+j].set(k);}else if(py[k] > yy) {b1[i*n+j].set(k);//printf("u1-%d", k);} else {b2[i*n+j].set(k);//printf("d1-%d", k);}} else {if(px[k] == x[i]) {b1[i*n+j].set(k);b2[i*n+j].set(k);}else if(px[k] > x[i])  {b1[i*n+j].set(k);//printf("u2-%d", k);} else {b2[i*n+j].set(k);//printf("d2-%d", k);}}//puts("");}//printf("  %d %d %d %d\n", i, j, b1[i*n+j].count(), b2[i*n+j].count());}}bitset
      
        tmp(0);int ans = 0;for(int i = 0; i < n; i ++) {for(int j = i+1; j < n; j ++) {for(int k = j+1; k < n; k ++) {if(check(i, j, k)) {tmp = b1[i*n+j];//printf("UU1 ");//put(b1[i*n+j]);}else {tmp = b2[i*n+j];//put(b2[i*n+j]);}if(check(i, k, j)) {tmp &= b1[i*n+k];//printf("UU2 ");//put(b1[i*n+k]);}else {tmp &= b2[i*n+k];//put(b2[i*n+k]);}if(check(j, k, i)) {tmp &= b1[j*n+k];//printf("UU3 ");//put(b1[j*n+k]);}else {tmp &= b2[j*n+k];//put(b2[j*n+k]);}//printf("***%d %d %d %d\n", i, j, k, tmp.count());if(tmp.count() & 1) ans ++;}}}printf("Case %d: %d\n", ++cas, ans);for(int i = 0; i < n*n; i ++) {b1[i].reset();b2[i].reset();}}return 0;}/*3 10 00 100100 00 03 10 00 100100 050 503 10 00 100100 0100 03 10 00 100100 00 -14 40 00 100100 0-2 500 00 100100 0-1 50*/