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题意
问最少用几条从原点出发的抛物线
即 \(ax^2+bx=0\) 能消灭全部的pig
题解
我是用dfs写的
对于每一个猪,检测是否已被击落,否则 暂时单独 or 与其他单独的猪组成抛物线
调试记录
有一个地方回溯忘记打了
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#define maxn 20
#define INF 0x3f3f3f3f
using namespace std;
int T, n, ans, tot = 0, totp = 0;
double tx[maxn], ty[maxn], px[maxn], py[maxn], x[maxn], y[maxn];
bool equ(double a, double b){
if (fabs(a - b) < 1e-8) return true;
return false;
}
void del(int pos){
for (int j = pos; j < tot; j++) tx[j] = tx[j + 1], ty[j] = ty[j + 1];
tot--;
}
void insert(int pos, double x, double y){
for (int j = tot; j >= pos; j--) tx[j + 1] = tx[j], ty[j + 1] = ty[j];
tot++;
tx[pos] = x, ty[pos] = y;
}
bool pass(int dep){
for (int k = 1; k <= totp; k++)
if (equ(px[k] * x[dep] * x[dep] + py[k] * x[dep], y[dep])) return true;
return false;
}
void dfs(int dep){
// printf("%d %d\n", tot, totp);
if (tot + totp >= ans) return;
if (dep == n + 1){
ans = min(ans, tot + totp);
return;
}
if (pass(dep)){
dfs(dep + 1);
return;
}
double x1 = x[dep], y1 = y[dep], x2 = 0, y2 = 0;
for (int i = 1; i <= tot; i++){
x2 = tx[i], y2 = ty[i];
if (equ(x1, x2)) continue;
double a = (x2 * y1 - x1 * y2) / (x1 * x1 * x2 - x1 * x2 * x2);
double b = (x1 * x1 * y2 - x2 * x2 * y1) / (x1 * x2 * (x1 - x2));
if (a < 0){
px[++totp] = a, py[totp] = b;
double ttx = tx[i], tty = ty[i];
del(i);
dfs(dep + 1);
insert(i, ttx, tty);
px[totp] = 0, py[totp--] = 0;
}
}
tx[++tot] = x[dep], ty[tot] = y[dep];
dfs(dep + 1);
tx[tot] = 0, ty[tot--] = 0;
}
int main(){
scanf("%d", &T);
while (T--){
int opt;
scanf("%d%d", &n, &opt);
for (int i = 1; i <= n; i++) scanf("%lf%lf", &x[i], &y[i]);
ans = INF;
totp = 0; memset(px, 0, sizeof px), memset(py, 0, sizeof py);
tot = 0; memset(tx, 0, sizeof tx), memset(ty, 0, sizeof ty);
dfs(1);
printf("%d\n", ans);
}
return 0;
}