POJ1328 Radar Installation(贪心)

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 77816		Accepted: 17409

Description

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. 


 

Figure A Sample Input of Radar Installations

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. 

The input is terminated by a line containing pair of zeros 

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.
Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

好题！真是一道好题啊！这一道题的意思就是说，在x轴上方是海域，下方和x轴都属于陆地。现在给定一些海岛坐标，并且给出雷达可以搜索的长度，要求雷达尽可能覆盖海岛，并且所用雷达数目最少。这道题，一开始想真的觉得好困惑，这咋整？后来慢慢形成思路，我把他们全部按照x轴从小到大排序，从最左边开始贪心，在雷达包含选定的海岛的情况下，尽可能放入更多的海岛。好，这是一个初始方案，接下来写起来就没这么简单咯。对了，雷达肯定是在x轴上的，这样照到的海域肯定会比在x轴下面要大！

问题1，选定第一个海岛之后，他在x轴上有两个交点，然后如果放入第二个海岛，这个交点的范围会一直缩小的，直到其他全部都不能放入这个交点范围内。而我一开始却只保留了右边的交点，因为我是以x轴从小到大排序的嘛，天真的以为只要下一个海岛左交点小于这个右交点就行了，但是发现我左边的交点也是变化的，也需要算出来才行。

问题2，排序的时候还要注意，如果x轴是相同的，那就按照y的大小从大到小排序，这样最上面那个点和x轴的两个交点的范围是相对于下面的点来说是比较窄的，可以减少一下范围的更新次数嘛。。。

问题3，我也是看到别人给的数据才想的，那就是如果真的这么奇葩他的海岛给的y点是小于0的，也就是在陆地上的怎么办，然后就是我一开始是用的m+n来结束while读入循环，然后发现他给我出了这样一组数据，3 -3，三个岛，雷达范围是-3？什么鬼？反正这样也会让我的程序结束，这样的数据我觉得是不合法的，但是他给出来了就改一改吧，毕竟acm本来就是个坑。

然后我自己还有各种各样的粗心大意的错误，TLE了两发，while循环那忘记要j++了。。。RE的两发，这个。。。我也不知道为什么。。。wa了一发，还是太粗心。。。然后，终于终于过了！

代码如下：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int mark[1005];
int n, m, ppp;
struct node
{
int x, y;
double left,right;
bool operator < (const node &a)const
{
if(x != a.x)
return x < a.x;
else
return y > a.y;
}
}island[1005];

void func(int i)
{
ppp++;
mark[i] = 1;
double LEFT = island[i].left, RIGHT = island[i].right;
int j = i + 1;
while(j < n)
{
if(!mark[j] && ((island[j].left >= LEFT && island[j].left <= RIGHT)
|| (island[j].right >= LEFT && island[j].right <= RIGHT)
|| (island[j].left <= LEFT && island[j].right >= RIGHT)))
{
mark[j] = 1;
LEFT = island[j].left > LEFT ? island[j].left : LEFT;
RIGHT = island[j].right < RIGHT ? island[j].right : RIGHT;
}
j++;
}
}

int main()
{
int flag, k = 1, i, j;
int tx, ty;
while(scanf("%d%d", &n, &m), n != 0 || m != 0)
{
memset(mark, 0, sizeof(mark));
flag = 0;
ppp = 0;
for(i = 0, j = 0; i < n; i++)
{
scanf("%d%d", &tx, &ty);
if(ty > m)
flag = 1;
if(ty >= 0 && !flag)
{
island[j].x = tx;
island[j].y = ty;
island[j].left = 1.0 * tx - sqrt(m * m - ty * ty);
island[j].right = 1.0 * tx + sqrt(m * m - ty * ty);
j++;
}
}
n = j;
printf("Case %d: ", k++);
if(flag || m <= 0)
printf("-1\n");
else
{
sort(island, island + n);
for(i = 0; i < n; i++)
{
if(!mark[i])
func(i);
}
printf("%d\n", ppp);
}
}
return 0;
}

看到代码才想起，就是我们判定下一个海岛是否在这个范围内，有三种情况，一种是，下一个海岛的左交点坐标在范围内，一种是右交点在范围内，还有一种！！！就是下一个海岛的左交点小于范围的左交点，右交点大于范围的右交点！相当于下一个海岛的左右范围比原本的范围还要大！这也是有交集的！！！脑子不够用我当时就是少了这种情况啊！！！好像就是因为这个wa了一发。

一年多了，重写了一次，发现还是不会。。。不过代码风格变得更清晰了

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cmath>
#include<utility>
#include<vector>
using namespace std;
#define eps 1e-7
typedef pair<double, double> pii;
vector <pii> v;
int n, d;

int main() {
int Case = 1, x, y;
while(cin >> n >> d) {

bool flag = 1;
v.clear();
if(n == 0 && d == 0)
break;
if(d < 0)
flag = 0;
for(int i = 0; i < n; i++) {
cin >> x >> y;

if(y <= 0)
continue;
if(y > d)
flag = 0;

if(!flag)
continue;

double tmp = sqrt((double)d * d - (double)y * y);
v.push_back(pii(x - tmp, x + tmp));
}

if(flag) {
int ans = 0;
sort(v.begin(), v.end());
for(int i = 0; i < v.size(); i++) {
double l = v[i].first;
double r = v[i].second;
int k = i + 1;
while(k < v.size() && v[k].second >= l && v[k].first <= r) {
l = max(v[k].first, l);
r = min(v[k].second, r);
k++;
}
i = k - 1;
ans++;
}
printf("Case %d: %d\n", Case++, ans);
} else {
printf("Case %d: -1\n", Case++);
}
}
return 0;
}