P1337 [JSOI2004] 平衡点 / 吊打XXX

合集 - 模拟(54)

1.G. Vlad and Trouble at MIT2024-02-202.P2899 [USACO08JAN] Cell Phone Network G2024-02-203.Watering an Array2024-02-234.P2055 [ZJOI2009] 假期的宿舍2024-03-085.T432579 大西洋船王2024-03-086.T432580 星云 hard ver.2024-03-087.P7072 [CSP-J2020] 直播获奖2024-03-098.E - Insert or Erase2024-03-109.D - String Bags2024-03-1010.P2671 [NOIP2015 普及组] 求和2024-03-1011.P1884 [USACO12FEB] Overplanting S2024-03-1012.F. Microcycle2024-03-1213.E. Increasing Subsequences__22024-03-1314.G. Path Prefixes2024-03-1415.F. Rudolf and Imbalance2024-03-1416.P1355 神秘大三角2024-03-2017.[ARC172A] Chocolate2024-03-2518.F. 0, 1, 2, Tree!2024-03-2919.E. Nearly Shortest Repeating Substring2024-03-2920.P1354 房间最短路问题2024-03-2921.P3258 [JLOI2014] 松鼠的新家2024-04-0222.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-05

23.P1337 [JSOI2004] 平衡点 / 吊打XXX2024-04-07

24.P3745 [六省联考 2017] 期末考试2024-04-0825.D. Inaccurate Subsequence Search2024-04-0926.E. Long Inversions2024-04-0927.C. Deep Down Below2024-04-0928.C. Rings2024-04-0929.F. Unfair Game2024-04-1030.C. Inversion Graph2024-04-1031.D. Cross Coloring2024-04-1232.D2. Set To Max (Hard Version)2024-04-1233.C. Array Game2024-04-1234.F. Alex's whims2024-04-1235.Queue Sort2024-04-1236.P10252 线性变换2024-04-1637.P10288 [GESP样题 八级] 区间2024-04-1638.P2161 [SHOI2009] 会场预约2024-04-1739.L2-001 紧急救援2024-04-1840.L2-002 链表去重2024-04-1841.L2-013 红色警报2024-04-1842.L2-014 列车调度2024-04-1943.L2-022 重排链表2024-04-1944.L2-023 图着色问题2024-04-1945.L2-039 清点代码库2024-04-1946.L2-034 口罩发放2024-04-1947.L3-037 夺宝大赛2024-04-2348.仓库规划2024-04-2349.因子化简2024-04-2350.C. Torn Lucky Ticket2024-04-2751.俄罗斯方块2024-05-1452.F. Cutting Game2024-05-2153.D. Prefix Permutation Sums2024-06-1754.P3304 [SDOI2013] 直径2024-07-15

收起

原题链接

题解

朝合力方向位移一段距离，并逐渐减小这个位移距离，需要痛苦的调参

code

#include<bits/stdc++.h>
using namespace std;
double x=0,y=0;
struct node
{
    double x,y,w;
}pos[1005];
int n;
void moves(double len)
{
    double fx=0,fy=0;
    for(int i=1;i<=n;i++)
    {
        double F=pos[i].w;
        double dis=0.000001+sqrt((pos[i].x-x)*(pos[i].x-x)+(pos[i].y-y)*(pos[i].y-y));//加1e-6是为了避免分母为零
        fx+=(pos[i].x-x)/dis*F;//计算x方向上合力
        fy+=(pos[i].y-y)/dis*F;//计算y方向上合力
    }

    x+=len*fx;//朝合力方向移动
    y+=len*fy;
}

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)  cin>>pos[i].x>>pos[i].y>>pos[i].w;

    double len=10000;

    while(len>0.000001)
    {
        moves(len);//振幅？模拟？
        len*=0.99;
    }

    printf("%.3lf %.3lf",x,y);
    return 0;
}