二分三角形的时候尤其需要注意！！！ HDU 5115 二分+模拟

Posted 2020-08-14 不知姓名的黑猫君

 

题目大意：http://blog.csdn.net/snowy_smile/article/details/49535301

思路：分类讨论，分别在[1,2]、（2,3）、[3,4）、[4,1]相遇，然后再特判v1和v2的大小关系即可。

然后特别需要注意在(2,3)时候的二分了，因为我们二分非斜边的话，非斜边的增长速率大于斜边的增长速率

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\\n")
const double eps = 1e-6;
const double len = 300.0;
const double duijiao = sqrt(1.0*300*300*2);
double t, v1, v2;
int sign(double x){
    return abs(x) < eps ? 0 : (x > 0 ? 1 : -1);
}

bool work_2_3(){
    double t1, t2;
    double lb = 0.0, rb = 300.0;
    for (int i = 1; i <= 100; i++){
        double mid = (lb + rb) / 2.0;
        t1 = sqrt(1.0*300*300 + mid*mid) / v1;
        t2 = (300.0 + mid) / v2;
        ///if (sign(t1 - t2) > 0) rb = mid;
        ///else lb = mid;
        if (sign(t1 - t2) > 0) lb = mid;
        else rb = mid;
    }
    t2 = (len - lb + len) / v2 + t;
    t1 = (lb + len * 2.0) / v1;
    if (sign(t1 - t2) > 0) return false;
    return true;
}

bool work_3_4(){
    double t1, t2;
    double lb = 0.0, rb = 300.0;
    for (int i = 1; i <= 100; i++){
        double mid = (lb + rb) / 2.0;
        t1 = (sqrt(1.0*300*300 + mid * mid)) / v1;
        t2 = (len * 3.0 - mid) / v2;
        if (sign(t1 - t2) > 0) rb = mid;
        else lb = mid;
    }
    t2 = (len + lb) / v2 + t;
    t1 = (sqrt(len*len + (len-lb)*(len-lb)) + 3*len) / v1;
    if (sign(t1 - t2) > 0) return false;
    return true;
}

bool solve(){
    ///[1,2]
    if (sign(v1 - v2) >= 0) return true;
    if (v1 == 0) return false;
    ///[1,4]
    double t1 = len / v1, t2 = len * 3.0 / v2;
    if (sign(t1 - t2) >= 0) return false;

    t1 = duijiao / v1;
    t2 = len * 2.0 / v2;
    if (sign(t1 - t2) >= 0) return work_3_4();
    else return work_2_3();
}

int main(){
    int kase; cin >> kase;
    for (int i = 1; i <= kase; i++) {
        scanf("%lf%lf%lf", &t, &v1, &v2);
        bool flag = solve();
        if (flag) printf("Case #%d: Yes\\n", i);
        else printf("Case #%d: No\\n", i);
    }
    return 0;
}

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