Radar Installation(POJ 1328)(贪心)
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题目链接:http://poj.org/problem?id=1328
Radar Installation
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 80287 Accepted: 17933
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
题意:
一个海上有n个小岛,现在要在海岸线,也就是x轴上布置雷达,每个雷达的信号范围为一个以雷达位置为圆心半径为d的圆,问最少需要多少雷达,才能把所有的岛全都覆盖到。岛的位置按直角坐标表示。
题解:
用贪心的思想求解。看看每个岛被覆盖时相应的雷达范围是什么,然后把范围按照范围左端大小排序。这时,就先覆盖最左边的那个范围,依次往右边的范围走。
1、如果现在扫描到的范围的左端点,比现在雷达所能放置的最右位置要大,那么就说明要放另一个雷达,并把雷达放置的最右位置更新为当前范围的右端点。
2、如果现在扫描到的范围的右端点,比现在雷达所能放置的最右位置要小,那么就说明这个雷达的最右范围要以这个比较小的位置为准,雷达放置的最右位置就要更新为这个较小值。
思考:
1、千万不能在边输入边处理的时候,处理到某个条件就直接退出循环,这样会使后面的数据读不进来,程序出现问题。
2、可以逆向思维,不想雷达布置在什么位置覆盖了什么点,而是想岛能接收到信号的雷达位置范围,也就是以岛为圆心,d为半径做圆,与x轴交点之间的范围内布置雷达,就能把这个岛覆盖到。
#include <iostream>#include <cstring>#include <cmath>#include <cstdio>#include <algorithm>using namespace std;struct Point{ double l, r;};const int maxn = 2010;int n;double d;int minn;Point p[maxn];int Compare (Point a, Point b){ return a.l < b.l;}int Calc(double x, double y, int i){ if (d < fabs(y)) return 0;//要取绝对值 double ans = sqrt (d * d - y * y); p[i].l = x * 1.0 - ans; p[i].r = x * 1.0 + ans; return 1;}void Count(){ sort (p, p + n, Compare); minn = 1; double Right = p[0].r; for (int i = 1; i < n; i++) { if (p[i].l > Right) { Right = p[i].r; minn++; } else if (p[i].r < Right) {//这样的话,肯定是包含关系,所以要把小的置为Right Right = p[i].r; } }}int main(){ #ifndef ONLINE_JUDGE freopen ("in.txt", "r", stdin); #endif // ONLINE_JUDGE int kase = 1; while (scanf ("%d%lf", &n, &d) != EOF) { if (n == 0 && d == 0) break; printf ("Case %d: ", kase++); int flag = 0;//如果不能搭建,则flag为1 for (int i = 0; i < n; i++) { double x, y; scanf ("%lf%lf", &x, &y); if (!Calc (x, y, i)) {//如果有一个岛无论怎样都没法被覆盖,则flag变为1 flag = 1;//不能在这里退出出去,因为在边输入边处理,如果从这里退出去,则后面的无法被读取!!! } } if (flag) { printf ("-1\n"); continue; } if (n == 0) { printf ("0\n"); continue; } Count (); printf ("%d\n", minn); } return 0;}
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