UVALive 6835

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抛物线过障碍

题目描述：

n~10个障碍物，一个球从a要求弹到b，然后中间一共弹小于c~15次，重力g=1，问可以越过障碍物最小的初速度是多少？

题解：

关键是枚举判定。先枚举几次，定出了两点，然后想出擦边走最好，那么枚举擦哪一个点，然后判定是否可以选最小的。但是记住45度是最好的，当障碍物太低的时候没必要擦边，45度走

重点：

45度走最好。枚举判定

代码：

//轩哥的代码

#include <cstdio>#include <algorithm>#include <cmath>#define EPS 1e-9using namespace std;double p[32], h[32];bool flag;int main() {    int n, b, doub, doubPre;    double d, dist, mul, ans;    while (~scanf("%lf%d%d", &d, &n, &b)) {        for (int i = 0; i < n; ++i)            scanf("%lf%lf", &p[i], &h[i]);        ans = -1;        for (int i = 0; i <= b; ++i) {//首先枚举跳几次，可以获得两个点            dist = d / (i+1);//45度的一定要特判。因为不在擦边走的范围内            mul = -1.0/dist;            flag = true;            for (int t = 0; t < n; ++t) {                doub = (int) ((p[t] / dist)+EPS);                //printf("pre---h[t]%f\n", mul*(p[t]- doub*dist)*(p[t] - (doub+1)*dist));                if (mul*(p[t]- doub*dist)*(p[t] - (doub+1)*dist) < h[t] - EPS) {                    //printf("h can't pass------%d\n", h[t]);                    flag = false;                    //printf("lll\n");                    break;                }            }            if (flag) {                if ((ans + 1.0) < EPS)                    ans = sqrt(-0.5/mul - mul*dist*dist/2.0);                else                    ans = min(ans,sqrt(-0.5/mul - mul*dist*dist/2.0));            }            else {//45度的不能，那么就只能擦边走了                for (int j = 0; j < n; ++j) {//枚举擦边谁，然后两根式确定抛物线，然后求出抛物线枚举判定                    doubPre = (int) ((p[j] / dist)+EPS);                    mul = h[j] / (p[j] - doubPre*dist) / (p[j] - (doubPre+1)*dist);                    flag = true;                    //printf("--- mul %f----dist %f----\n", mul, dist);                    for (int t = 0; t < n; ++t) {                        doub = (int) ((p[t] / dist)+EPS);                        //printf("beishu----%d\n", doub);                        //printf("-----h[t]%f\n", mul*(p[t]- doub*dist)*(p[t] - (doub+1)*dist));                        if (mul*(p[t]- doub*dist)*(p[t] - (doub+1)*dist) < h[t] - EPS) {                        //  printf("h------%d\n", h[t]);                            flag = false;                        //  printf("lll\n");                            break;                        }                    }                    if (flag) {                        //printf("okokok----- mul %f----dist %f----\n", mul, dist);                        //printf("vx---%f   vy----%f\n", sqrt(-0.5/mul), sqrt(-mul*dist*dist/2.0));                        if ((ans + 1.0) < EPS)                            ans = sqrt(-0.5/mul - mul*dist*dist/2.0);                        else                            ans = min(ans,sqrt(-0.5/mul - mul*dist*dist/2.0));                    }                }            }        }        printf("%f\n", ans);    }    return 0;}

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