簡單題 —— Crossing Rivers —— 2009 Asia Wuhan Regional Contest Hosted by Wuhan University

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3232

Crossing Rivers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 225

Problem Description

You live in a village but work in another village. You decided to follow the straight path between your house (A) and the working place (B), but there are several rivers you need to cross. Assume B is to the right of A, and all the rivers lie between them.

Fortunately, there is one "automatic" boat moving smoothly in each river. When you arrive the left bank of a river, just wait for the boat, then go with it. You're so slim that carrying you does not change the speed of any boat.

Days and days after, you came up with the following question: assume each boat is independently placed at random at time 0, what is the expected time to reach B from A? Your walking speed is always 1.

To be more precise, for a river of length L, the distance of the boat (which could be regarded as a mathematical point) to the left bank at time 0 is uniformly chosenfrom interval [0, L], and the boat is equally like to be moving left or right, if it’s not precisely at the river bank.

 

Input

There will be at most 10 test cases. Each case begins with two integers n and D, where n (0 <= n <= 10) is the number of rivers between A and B, D (1 <= D <= 1000) is the distance from A to B. Each of the following n lines describes a river with 3 integers: p, L and v (0 <= p < D, 0 < L <= D, 1 <= v <= 100). p is the distance from A to the left bank of this river, L is the length of this river, v is the speed of the boat on this river. It is guaranteed that rivers lie between A and B, and they don’t overlap. The last test case is followed by n=D=0, which should not be processed.

 

Output

For each test case, print the case number and the expected time, rounded to 3 digits after the decimal point.

Print a blank line after the output of each test case.

 

Sample Input

1 1 0 1 2 0 1 0 0

 

Sample Output

Case 1: 1.000 Case 2: 1.000

 

Source

2009 Asia Wuhan Regional Contest Hosted by Wuhan University

 

Recommend

chenrui

 ————————————————————————————————————————————————————————————————

          題目大意：說從A（左邊）到B（右邊），之間有些河流，每條河流有一條船。告訴你A、B之間的距離，河流的數量，河流距離A的距離，河流的寬度，船在這條河中的速度。然後是每條河中的船不一定在初始狀態時就在河邊，而是隨機的在河中央，或在河的對面。但這都不影響。因爲第一個樣例等於已經告訴你概率的計算公式了。

       做這題的唯一不明白的地方，爲什麼ans（即輸出的總時間）用float會wrong，而用double才AC，可能和精度有關吧，但只要求小數點後3位啊。不懂。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <string>
#include <bitset>

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>

#include <map>
#include <stack>
#include <queue>
#include <set>
#include <list>

#include <utility>

using namespace std;


struct boat
{
    int p, l, v;
    bool operator < (const struct boat bb)const
    {
        return this -> p - bb.p;
    }
  //  bool operator > (const struct boat bb)const
  //  {
 //       return this -> p - bb.p;
 //   }
};

struct boat b[500];


int cmp(const void *x, const void *y)
{
    return ((struct boat*)x) -> p - ((struct boat*)y) -> p;
}

int main()
{
//	srand(time(NULL));
    int n, d, lct,ca;
    double ans;
    ca = 1;
    scanf("%d %d", &n ,&d);
    while (  n != 0 || d != 0)
    {
        ans = 0.000000;
        for (int j = 0; j < n; j++)
        {
            cin >> b[j].p >> b[j].l >> b[j].v;
        }
        qsort(b, n, sizeof (struct boat), cmp);
       // sort(b, b + n);
        lct = 0;
        for (int j= 0; j < n; j++)
        {
            ans = ans + (b[j].p - lct);
            lct = b[j].p + b[j].l;
            ans += (float)(b[j].l * 2 + 0.0000) /(float)(0.0000+  b[j].v);

        }
        ans = ans + d - lct;
        printf("Case %d: %.3f\n\n",ca++, ans);
        scanf("%d %d", &n ,&d);
    }

	return 0;
}