P2831 愤怒的小鸟
从 \((0, 0)\) 发射一只鸟, 轨迹满足抛物线, 问最少几只鸟可以打完 \(n <= 18\) 只猪
错误日志: 处理抛物线数组没有初始化
Solution
数据范围识状压
挺好想的状压dp
对于每只猪考虑两种情况: 自己被单独打下来或者被其他抛物线经过打下来
现在已经有一个确定点 \((0 ,0)\) 再加一只猪, 两个点无法确定一条抛物线
也就是说这条抛物线是可以给你自己规划的
也也就是说自己被单独打下来一定可行
不过能打多一点会更优
三点确定抛物线
我们暴力枚举两只没被打下来的猪
看还能打几只
更新dp即可
注意两点:
预处理出所有抛物线能打几只猪, 可以省一维枚举每只猪, 不然 \(TLE\)
此题卡精度, 判断两个浮点数是否相等可以确定一个精度 (此题设置为 \(10^{-6}\))
Code
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#include<cmath>
#define LL long long
#define REP(i, x, y) for(int i = (x);i <= (y);i++)
using namespace std;
int RD(){
int out = 0,flag = 1;char c = getchar();
while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
return flag * out;
}
const int maxn = (1 << 19);
int T;
int num;
int dp[maxn], one[25];
double a, b;
inline void build(double &a,double &b,double x1,double y1,double x2,double y2){
a=(x2*y1-x1*y2)/(x1*x2*(x1-x2));
b=(x1*x1*y2-x2*x2*y1)/(x1*x2*(x1-x2));
}//计算a,b
bool judge(double x, double y){
double y1 = a * x * x + b * x;
if(fabs(y - y1) <= 1e-6)return 1;
return 0;
}
struct Node{
double x, y;
}I[25];
int van[25][25];
void init(){
memset(van, 0, sizeof(van));
REP(i, 1, num){
REP(j, i + 1, num){
if(fabs(I[i].x - I[j].x) <= 1e-6)continue;
build(a, b, I[i].x, I[i].y, I[j].x, I[j].y);
if(a >= 0)continue;
REP(k, 1, num){
if(judge(I[k].x, I[k].y))van[i][j] |= one[k];
}
}
}
}
int main(){
T = RD();
while(T--){
num = RD(), RD();
REP(i, 1, num)scanf("%lf %lf", &I[i].x, &I[i].y), one[i] = (1 << (i - 1));
int maxstate = (1 << num) - 1;
REP(i, 0, maxstate)dp[i] = 1e9;
init();
dp[0] = 0;
REP(i, 0, maxstate){//状态
REP(j, 1, num){//一只新猪
if(i & one[j])continue;
dp[i | one[j]] = min(dp[i | one[j]], dp[i] + 1);//看看只打一只猪赚不赚
REP(k, j + 1, num){//另一只新猪
if(i & one[k] || !van[j][k])continue;
dp[i | van[j][k]] = min(dp[i | van[j][k]], dp[i] + 1);
}
}
}
printf("%d\n", dp[maxstate]);
}
return 0;
}