acm 1011 雷达数目
2016-03-31 20:01
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1.1011
2.
[align=left]Problem Description[/align]
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
[align=left]Input[/align]
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. <br> <br>The input is terminated by a line containing pair of zeros <br>
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
3 2
1 2
-3 1
2 1
1 2
0 2
0 0
[align=left]Sample Output[/align]
Case 1: 2
Case 2: 1
3.求能够覆盖所有岛屿的最小雷达数目
4.每个小岛对应x轴上的一个区间,在这个区间内的任何一个点放置雷达,则可以覆盖该小岛,区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样,问题转化为已知一定数量的区间,求最小数量的点,使得每个区间内至少存在一个点。每次迭代对于第一个区间, 选择最右边一个点, 因为它可以让较多区间得到满足, 如果不选择第一个区间最右一个点(选择前面的点),
那么把它换成最右的点之后, 以前得到满足的区间, 现在仍然得到满足, 所以第一个区间的最右一个点为贪心选择, 选择该点之后, 将得到满足的区间删掉, 进行下一步迭代, 直到结束(参考杨世诚)
5.
#include<iostream>
#include<algorithm>
#include<numeric>
#include<math.h>
using namespacestd;
struct str
{
double l;
double r;
};
bool cmp(conststr&a,const str&b)
{
if(a.l<=b.l) return true;
return false;
}
int main()
{
str r[1000];
int n,d,i,l;
int x,y,t,k,e=1;
double a;
while(cin>>n>>d&&n!=0||d!=0)
{
t=1;l=1;
for(i=0;i<n;i++)
{
cin>>x>>y;
if(d>=y&&l)
{
r[i].l=x-sqrt(d*d-y*y);
r[i].r=x+sqrt(d*d-y*y);
}
else l=0;
}
if(!l)
{
cout<<"Case"<<e++<<": -1"<<endl;
continue;
}
sort(r,r+n,cmp);
a=r[0].r;
for(k=1;k<n;k++)
{
if(r[k].l>a)
{
t++;
a=r[k].r;
}
else if(r[k].r<=a)
{
a=r[k].r;
}
}
cout<<"Case"<<e<<": "<<t<<endl;
e++;
}
return 0;
}
2.
[align=left]Problem Description[/align]
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
[align=left]Input[/align]
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. <br> <br>The input is terminated by a line containing pair of zeros <br>
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
3 2
1 2
-3 1
2 1
1 2
0 2
0 0
[align=left]Sample Output[/align]
Case 1: 2
Case 2: 1
3.求能够覆盖所有岛屿的最小雷达数目
4.每个小岛对应x轴上的一个区间,在这个区间内的任何一个点放置雷达,则可以覆盖该小岛,区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样,问题转化为已知一定数量的区间,求最小数量的点,使得每个区间内至少存在一个点。每次迭代对于第一个区间, 选择最右边一个点, 因为它可以让较多区间得到满足, 如果不选择第一个区间最右一个点(选择前面的点),
那么把它换成最右的点之后, 以前得到满足的区间, 现在仍然得到满足, 所以第一个区间的最右一个点为贪心选择, 选择该点之后, 将得到满足的区间删掉, 进行下一步迭代, 直到结束(参考杨世诚)
5.
#include<iostream>
#include<algorithm>
#include<numeric>
#include<math.h>
using namespacestd;
struct str
{
double l;
double r;
};
bool cmp(conststr&a,const str&b)
{
if(a.l<=b.l) return true;
return false;
}
int main()
{
str r[1000];
int n,d,i,l;
int x,y,t,k,e=1;
double a;
while(cin>>n>>d&&n!=0||d!=0)
{
t=1;l=1;
for(i=0;i<n;i++)
{
cin>>x>>y;
if(d>=y&&l)
{
r[i].l=x-sqrt(d*d-y*y);
r[i].r=x+sqrt(d*d-y*y);
}
else l=0;
}
if(!l)
{
cout<<"Case"<<e++<<": -1"<<endl;
continue;
}
sort(r,r+n,cmp);
a=r[0].r;
for(k=1;k<n;k++)
{
if(r[k].l>a)
{
t++;
a=r[k].r;
}
else if(r[k].r<=a)
{
a=r[k].r;
}
}
cout<<"Case"<<e<<": "<<t<<endl;
e++;
}
return 0;
}
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