POJ 1328 -- Radar Installation


http://poj.org/problem?id=1328
Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.
Input
The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 
3 2
1 2
-3 1
2 1
http://blog.csdn.net/petercsj/article/details/4602946
贪心算法:我们先将问题的区间集合按右端点排序,从最左边的区间开始,如果该区间内雷达为空为空,则选区间最右边的端点放雷达,并删掉左端点在雷达前的区间,进入子问题。
下面来证明算法的正确性:
即要证两点:
(1)贪心选择性
(2)最优子结构性质
 首先我们先来想想最原始的做法。就是在一堆区间中选几个点。那么我们可以拆分成若干步,每一步选一个点。选过之后,包含该点的区间都被覆盖,这样原问题划分为两个子问题,并显然满足最优子结构性质。假如雷达的坐标限定为整数,那么我们就可以采用动态规划的方法来解题。但雷达的坐标是实数,所以我们只有另辟蹊径。
经验,应该来说是通过经验,我们想到了上文所说得贪心算法。至于为什么这么做,只能说贪心算法的模式一般都是在极端处取值(可以参照活动选择问题的解法)。如果我们知道了该选择是贪心选择,即最优解的一种包含该选择,那么我们再根据上文叙述的最优子结构性质即可证明算法的正确性。(即求得的解不会比包含贪心选择的最优解大,否则与最优子问题矛盾)
    下面我们证明该选择是贪心选择。
首先我们证明,子问题的第一个区间必定只能包含一个点。因为在求解子问题之前定下的点不会落在该区间,否则该区间就不会出现在子问题中;并且在求解该子问题时,该区间也不会落入两个点,否则我们去掉左边那个点,对问题没有影响(本质是因为按尾部递增排列)。请看图示:
   ..
 ----
  ----
   ----
 
    每一个点“照看”一簇区间,左边那个点能照看到的区间,右边那个点都能照看到。因此在最优解中,第一个区间有且只有一个点。然而其他的区间则不一定,因为在子问题中新增的点可能落在之前被删掉的区间内。
    这点明确之后,我们就可以开始我们的证明。因为第一个区间有且只有一点,那么假设在最优解中这一点不是放在最右端点,那么我们可以把这一点移到最右端点,并不影响它覆盖的islands。原因就不必赘述了。这样我们就得到了另一个最优解。因此,选第一个区间的最右端点是安全的选择,亦即贪心选择。
    这样我们就比较完整的证明了算法的正确性了。

Code from http://www.hankcs.com/program/cpp/poj-1328-radar-installation-challenge-programming-contest-2nd-edition-exercises-answers.html
贪心策略依然是从左往右,尽量让每颗雷达覆盖最大岛屿数。
对整个题目数据的处理有个思维转换的过程,如果从雷达这个点出发去思考,那么会陷入double无尽的遍历中去。
我的思路是,先对每个海岛求一个区间:能覆盖它的所有雷达的圆心所构成的区间。然后对区间排序,定义一个最右点end,依次延伸end,如果end不在某个区间内,那么消耗一颗雷达,end更新为该区间的最右端,否则end更新为起点在end内的所有区间的最小右端点。
struct Section
{
    double begin;
    double end;
    bool operator < (const Section& b) const 
    {  
        return begin < b.begin;
    
};
int main(int argc, char *argv[])
{
#ifndef ONLINE_JUDGE
    freopen("in.txt""r", stdin);
    freopen("out.txt""w", stdout);
#endif
    int n, d, __id;
    __id = 0;
    while(cin >> n >> d && n > 0)
    {
        int result = 0;
        vector<Section> s(n);
        for (int i = 0; i < n; ++i)
        {
            double x, y;
            cin >> x >> y;
            if (result == -1)
            {
                continue;
            }
            if (y > d)
            {
                // 如果y比半径大,一定不能覆盖
                result = -1;
                continue;
            }
            double r = sqrt(d * d - y * y);
            s[i].begin = x - r;
            s[i].end = x + r;
        }
 
        if (result == -1)
        {
            cout << "Case " << ++__id << ": " << result << endl;
            continue;
        }
 
        sort(s.begin(), s.end());
        double end = -numeric_limits<double>::max();
        for (vector<Section>::iterator it = s.begin(); it != s.end(); ++it)
        {
            // cout << it->begin << " " << it->end << endl;
            if (end < it->begin)
            {
                ++result;
                end = it->end;
            }
            else if (end > it->end)
            {
                end = it->end;
            }
        }
 
        cout << "Case " << ++__id << ": " << result << endl;
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("out.txt");
#endif
    return 0;
}

Also check http://ren.iteye.com/blog/344093
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