题意：有$n$个小学生需要到距离为$l$的地方去，步行的速度是$v_1$，它们租了一辆大巴，速度是$v_2$，大巴上最多容纳$k$个乘客，每个小学生最多乘车一次，初始时大巴和小学生都在起点，问至少需要多长时间所有小学生都可以到达终点。时间精确到$10^{-6}$。数据范围$1 \leq l \leq 1e9, 1 \leq v_1 < v_2 \leq 1e9, 1 \leq k \leq n$。
分析：你可能会想到用大巴把小学生送到终点再折回，如此反复，直到最后一次不折回。但是样例$2$提示我们这种贪心不是最优的，大巴不一定要把小学生送到终点，因此考虑二分时间，那么假设时间$t$内可以完成目标，可以导出每个小学生的乘车距离最少是$\frac{v_2(l-v_1t)}{v_2-v_1}$，并且大巴需要折回至少$\frac{n}{ k }- [n \mod k = 0]$次，于是根据这两个信息得出最小花费时间与$t$作比较即可完成二分。代码如下：
1 #include <algorithm>
2 #include <cstdio>
3 #include <cstring>
4 #include <string>
5 #include <queue>
6 #include <map>
7 #include <set>
8 #include <stack>
9 #include <ctime>
10 #include <cmath>
11 #include <iostream>
12 #include <assert.h>
13 #pragma comment(linker, "/STACK:102400000,102400000")
14 #define max(a, b) ((a) > (b) ? (a) : (b))
15 #define min(a, b) ((a) < (b) ? (a) : (b))
16 #define mp std :: make_pair
17 #define st first
18 #define nd second
19 #define keyn (root->ch[1]->ch[0])
20 #define lson (u << 1)
21 #define rson (u << 1 | 1)
22 #define pii std :: pair<int, int>
23 #define pll pair<ll, ll>
24 #define pb push_back
25 #define type(x) __typeof(x.begin())
26 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
27 #define FOR(i, s, t) for(int i = (s); i <= (t); i++)
28 #define ROF(i, t, s) for(int i = (t); i >= (s); i--)
29 #define dbg(x) std::cout << x << std::endl
30 #define dbg2(x, y) std::cout << x << " " << y << std::endl
31 #define clr(x, i) memset(x, (i), sizeof(x))
32 #define maximize(x, y) x = max((x), (y))
33 #define minimize(x, y) x = min((x), (y))
34 using namespace std;
35 typedef long long ll;
36 const int int_inf = 0x3f3f3f3f;
37 const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
38 const int INT_INF = (int)((1ll << 31) - 1);
39 const double double_inf = 1e30;
40 const double eps = 1e-14;
41 typedef unsigned long long ul;
42 typedef unsigned int ui;
43 inline int readint(){
44 int x;
45 scanf("%d", &x);
46 return x;
47 }
48 inline int readstr(char *s){
49 scanf("%s", s);
50 return strlen(s);
51 }
52
53 class cmpt{
54 public:
55 bool operator () (const int &x, const int &y) const{
56 return x > y;
57 }
58 };
59
60 int Rand(int x, int o){
61 //if o set, return [1, x], else return [0, x - 1]
62 if(!x) return 0;
63 int tem = (int)((double)rand() / RAND_MAX * x) % x;
64 return o ? tem + 1 : tem;
65 }
66 ll ll_rand(ll x, int o){
67 if(!x) return 0;
68 ll tem = (ll)((double)rand() / RAND_MAX * x) % x;
69 return o ? tem + 1 : tem;
70 }
71
72 void data_gen(){
73 srand(time(0));
74 freopen("in.txt", "w", stdout);
75 int kases = 40000;
76 printf("%d\n", kases);
77 while(kases--){
78 ll sz = 1e18;
79 printf("%lld\n", Rand(sz, 1));
80 }
81 }
82
83 struct cmpx{
84 bool operator () (int x, int y) { return x > y; }
85 };
86 double power(double a, int p){
87 double ans = 1.;
88 while(p){
89 if(p & 1) ans *= a;
90 p >>= 1;
91 a = a * a;
92 }
93 return ans;
94 }
95 int main(){
96 //data_gen(); return 0;
97 //C(); return 0;
98 int debug = 0;
99 if(debug) freopen("in.txt", "r", stdin);
100 //freopen("out.txt", "w", stdout);
101 int n, l, v1, v2, k;
102 while(~scanf("%d", &n)){
103 l = readint(), v1 = readint(), v2 = readint(), k = readint();
104 int lamda = n / k - (n % k == 0);
105 double L = (double)l / v2, R = (double)l / v1;
106 while(R - L > 1e-6){
107 //printf("%.10f %.10f\n", L, R);
108 //printf("err is %.10f\n", R - L);
109 double mid = (L + R) / 2;
110 //printf("mid is %.10f\n", mid);
111 double d = (double)v2 * (l - v1 * mid) / (v2 - v1);
112 double lhs = 2. * d * v1 * lamda / (v1 + v2) + d;
113 int ok = 1;
114 if(lhs >= l) ok = 0;
115 if(ok){
116 double rhs = 2. * d * lamda / (v1 + v2);
117 rhs += (l - lhs + d) / v2;
118 if(rhs > mid) ok = 0;
119 }
120 if(ok) R = mid;
121 else L = mid;
122 }
123 //if(n % k) ans += d / v2;
124 printf("%.10f\n", L);
125 }
126 return 0;
127 }
code：
这里还需要注意二分的精度不能太高，否则可能由于浮点误差跳不出循环。
时间： 2024-11-08 23:20:57