技術標籤：POJ演算法c#c++

題目描述

總時間限制:1000ms 記憶體限制:65536kB

描述

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

輸入

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

輸出

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.

樣例輸入

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

樣例輸出

Case 1: 2
Case 2: 1

問題解決

沒做作業前聽了老師講的思路，就一脈相承的做了這個思路的解決方法QAQ

一個比較巧妙的地方是，在輸入資料的時候，首先將其座標轉化為在x軸上的一個區間，也就是說雷達要覆蓋這個島，就只能裝在這個範圍內。

然後將問題轉化為【點對區間的覆蓋】的問題，使用貪心法解決，將所有區間按起點排序，然後遍歷一遍，每遇到一個新的區間，分為兩種情況：

1.下一個區間完全在現有區間之外——結果加一，現有區間更新為下一個區間

2.下一個區間和現有區間有交集——將現有區間更新為現有區間與該區間的交集

#include <iostream>
#include <algorithm>
using namespace std;

struct range{
    double l,r;
}ranges[1005];

bool cmp(range x,range y){
    return x.l<y.l;
}

int main(){
    int n,d,cnt=1;
    while(cin>>n>>d){
        if (n==0 & d==0) break;
        int x,y,res=1;
        for(int i=0;i<n;i++){
            cin>>x>>y;
            if(y>d) res=-1;
            else{
                ranges[i].l=x-sqrt(d*d-y*y);
                ranges[i].r=x+sqrt(d*d-y*y);
            }
        }
        if(res==-1){
            cout<<"Case "<<cnt++<<": "<<res<<endl;
            continue;
        }
        sort(ranges,ranges+n,cmp);
        double start=ranges[0].l,end=ranges[0].r;
        for(int i=1;i<n;i++){
            if(ranges[i].l>end){
                start=ranges[i].l;
                end=ranges[i].r;
                res++; 
            }else {
                start=ranges[i].l;
                end=min(ranges[i].r,end);
            }
        }
        cout<<"Case "<<cnt++<<": "<<res<<endl;
    }
    return 0;
}

最最最坑的地方是，千萬不能給變數int！比如我的start和end一不小心給了int，找了好久才找出來錯TUT