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Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 42461 | Accepted: 9409 |
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
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 21 2-3 12 11 20 20 0
Sample Output
Case 1: 2Case 2: 1
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 using namespace std; 9 10 struct node11 {12 double start;13 double end;14 }coor[1001];//记录每个区间的端点15 int cmp(const struct node a,const struct node b)16 {17 return a.start < b.start;18 }19 int t,r;20 stack st;//用栈存每个区间，21 22 int cal(int ans)23 {24 while(!st.empty())25 st.pop();26 for(int i = t-1; i >= 0; i--)27 st.push(coor[i]);28 while(st.size() >= 2)//当栈中至少存在两个区间时29 {30 struct node tmp1 = st.top();31 st.pop(); 32 struct node tmp2 = st.top();33 if(tmp1.end >= tmp2.start)//当取出的两个区间有公共部分时34 {35 st.pop();//tmp2出栈36 struct node tmp;37 tmp.start = max(tmp1.start, tmp2.start);//注意取公共部分时，起始点取较大者38 tmp.end = min(tmp1.end, tmp2.end);//终点取较小者39 st.push(tmp);//将公共部分入栈40 ans--;//每两个区间交一次，雷达个数减一次41 }42 }43 return ans;44 }45 int main()46 {47 48 int cor_x[1001],cor_y[1001];49 double add;50 int cnt = 1;51 while(~scanf("%d %d",&t,&r))52 {53 int ok = 1;//判断小岛的坐标是否合法，54 if(t == 0 && r == 0) break;55 for(int i = 0; i < t; i++)56 {57 scanf("%d %d",&cor_x[i],&cor_y[i]);58 if(cor_y[i] > r)//若小岛纵坐标大于半径则不合法59 ok = 0;60 }61 if(ok == 0)62 {63 printf("Case %d: -1\n", cnt++);64 continue;65 }66 for(int i = 0; i < t; i++)67 {68 //以每个小岛为圆心，r为半径画圆，coor[]存该圆与x轴相交的区间69 add = sqrt(r*r-cor_y[i]*cor_y[i]);70 coor[i].start = cor_x[i] - add;71 coor[i].end = cor_x[i] + add;72 }73 sort(coor,coor+t,cmp);//对这些区间按起始点从小到大排序74 int ans = t;75 ans = cal(ans);76 printf("Case %d: %d\n",cnt++,ans);77 }78 return 0;79 }