HDU 4816 Bathysphere（数学）（2013 Asia Regional Changchun）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4816

Problem Description

The Bathysphere is a spherical deep-sea submersible which was unpowered and lowered into the ocean on a cable, and was used to conduct a series of dives under the sea. The Bathysphere was designed for studying undersea wildlife.

The Bathysphere was conducted from the deck of a ship. After counted, the ship should not move, so choosing the position where the Bathysphere was conducted is important.

A group of scientists want to study the secrets of undersea world along the equator, and they would like to use the Bathysphere. They want to choose the position where the Bathysphere can dive as deep as possible. Before conducting the Bathysphere, they have a map of the seabed, which tell them the shape of the seabed. They draw a line on the equator of the map to mark where they will release the Bathysphere, as a number axis. Suppose the axis is draw from 0 to L. But when they release the Bathysphere, they can't know where they are accurately, i.e., if they choose position x to release the Bathysphere, the real position will distribute between x-d and x+d with an equal probability, where d is given. The objective of the scientists is very simple, i.e., to maximize the expected depth.

For the ease of presentation, the shape of the seabed is described as a poly line. Given N points ) , ( X_i,Y_i ) as the vertices, where X_i and Y_i indicate the position and the depth of the i-th vertex, respectively, the ploy line is composed of the line segments that connect consecutive vertices.

 

Input

The first line contains an integer T (1 ≤ T ≤ 25), the number of test cases.

Then T test cases follow. In each test case, the first line contains two integers N (2 ≤ N ≤ 2*10^5) and L (2 ≤ L ≤ 10^9), as described above. Then N lines follow, each line contains two integer X_i and Y_i (1≤i≤N, 0≤ Y_i ≤10^9), where point ( X_i,Y_i ) is a vertex of the ploy line. It is assumed that X₁ == 0 and X_n == L and X_i < X_i+1 for 1 ≤ i < N. Then the following line contains one integer d (0 ≤ d ≤ L/2), as described above.

 

Output

For each test case, choose a position between d and L-d, both inclusive, to conducted the Bathysphere, and calculate the expected depth. Output the expected depth in a line, rounded to 3 digits after the decimal point.

 

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贴一下官方题解：

题目大意：
在海平面上找一点投放潜水艇，投放的准确地点存在误差D，求最大的潜水深度期望。
题目分析：
即在海平面下再画一条折线，然后用间距为2×D的竖线将图截出，求截出的图形的最大面积。
解法：
可以看出当将两竖线不断右移的过程中，除了一种状态以外，其余状态对于面积的影响均为单调的。
此状态为当左边竖线所相交的折线为向上趋势并且右边竖线所相交的折线为向下趋势并且在到达端点前，两竖线与折线的交点的高度为相同的值时，此时面积最大。
所以，可以直接将两竖线从左往右移动，每次移动一个端点的距离，如果出现该情况则计算中途可能出现的最大面积，否则记录当前最大面积，即可于O(N)时间内得出结果。

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PS：我写这题在HDU上不用long double就过不了。也完全不知道怎么用double过了，有知道怎么办的请务必告诉我！三分就不要了……

代码（2390MS）：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 typedef long double LDB;

 const int MAXN = ;
 const LDB EPS = 1e-;

 inline int sgn(LDB x) {
     return (x > EPS) - (x < -EPS);
 }

 int x[MAXN], y[MAXN];
 int n, d, L, T;

 struct Game {
     LDB a, b, c;
     Game() {}
     Game(LDB a, LDB b, LDB c): a(a), b(b), c(c) {}
     Game operator - (const Game &rhs) const {
         return Game(a - rhs.a, b - rhs.b, c - rhs.c);
     }
     LDB val_at(LDB x) {
         return a * x * x + b * x + c;
     }
     LDB max_val(LDB l, LDB r) {
         LDB res = max(val_at(l), val_at(r));
         if(sgn(a) < ) {
             LDB t = - b /  / a;
             if(sgn(l - t) <=  && sgn(t - r) <= )
                 res = val_at(t);
         }
         return res;
     }
 };

 Game get(int pos, LDB v = 0.0) {
     LDB k = LDB(y[pos + ] - y[pos]) / (x[pos + ] - x[pos]);
     LDB t = y[pos] - k * x[pos];
     LDB a = k / , b = t, c = -((k / ) * x[pos] + t) * x[pos];
     return Game(a,  * v * a + b, a * v * v + b * v + c);
 }

 LDB area(int pos) {
     return (y[pos] + y[pos + ]) / 2.0 * (x[pos + ] - x[pos]);
 }

 LDB solve() {
     if(d == ) {
         int res = ;
         for(int i = ; i <= n; ++i) res = max(res, y[i]);
         return res;
     }
     LDB nowx = , res = , s = ;
     int l = , r = ;
     while(r < n && x[r + ] <= d)
         s += area(r++);
     if(r == n) res = s;
     while(r < n) {
         LDB minx = min(x[l + ] - nowx, x[r + ] - nowx - d);
         res = max(res, s + (get(r, d) - get(l)).max_val(nowx, nowx + minx));
         nowx += minx;
         if(sgn(x[l + ] - nowx) == ) s -= area(l++), nowx = x[l];
         if(sgn(x[r + ] - nowx - d) == ) s += area(r++), nowx = x[r] - d;
     }
     return res / d;
 }

 int main() {
     scanf("%d", &T);
     while(T--) {
         scanf("%d%d", &n, &L);
         for(int i = ; i <= n; ++i) scanf("%d%d", &x[i], &y[i]);
         x[n + ] = x[n];
         scanf("%d", &d); d <<= ;
         printf("%.3f\n", (double)solve());
     }
 }