算法设计与分析 模拟考 radar installation

总时间限制: 

1000ms 

内存限制: 

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

输入

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

输出

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.

样例输入

3 21 2-3 12 11 20 20 0

样例输出

Case 1: 2Case 2: 1

来源

Beijing 2002

计算每个点对应的圆心区间范围，之后贪心即可。

注意采用优先队列的时候需要使用结构体对于小于进行重载。

采用优先队列的写法：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <vector>#include <algorithm>#include <queue>#include <math.h>using namespace std;int n;int d;int casenum;typedef struct ra{    float l;    float r;    bool operator < (const ra& b) const    {        if(l == b.l) return r > b.r;        else            return l > b.l;    }}range;priority_queue<ra> pq;void getradis(int x, int y){    range a;    float t = sqrt(d * d - y * y);    a.l = x - t;    a.r = x + t;    pq.push(a);}int getnum(){    int num = 0;    while(!pq.empty())    {        num++;        float radis = pq.top().r;        pq.pop();        while(!pq.empty())        {            range next = pq.top();            if(next.l <= radis)            {                if(next.r >= radis)                {                    pq.pop();                    continue;                }                else                {                    radis = next.r;                    pq.pop();                    continue;                }            }            else            {                break;            }        }    }    return num;}int main(){    scanf("%d%d", &n, &d);    while(n != 0 || d != 0)    {        casenum++;        while(!pq.empty())            pq.pop();        bool flag = true;        for(int i = 0; i < n; i++)        {            int x, y;            scanf("%d%d", &x, &y);            if(y <= d)                getradis(x, y);            else                flag = false;        }        if(flag)            printf("Case %d: %d\n", casenum, getnum());        else            printf("Case %d: %d\n", casenum, -1);        scanf("%d%d", &n, &d);    }    return 0;}

采用sort的写法：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <vector>#include <algorithm>#include <queue>#include <math.h>using namespace std;int n;int d;int casenum;/*typedef struct ra{    float l;    float r;    bool operator < (const ra& b) const    {        if(l == b.l) return r > b.r;        else            return l > b.l;    }}range;priority_queue<ra> pq;void getradis(int x, int y){    range a;    float t = sqrt(d * d - y * y);    a.l = x - t;    a.r = x + t;    pq.push(a);}int getnum(){    int num = 0;    while(!pq.empty())    {        num++;        float radis = pq.top().r;        pq.pop();        while(!pq.empty())        {            range next = pq.top();            if(next.l <= radis)            {                if(next.r >= radis)                {                    pq.pop();                    continue;                }                else                {                    radis = next.r;                    pq.pop();                    continue;                }            }            else            {                break;            }        }    }    return num;}*/vector< vector<float> > pq;void getradis(int x, int y){    vector<float> a;    float t = sqrt(d * d - y * y);    a.push_back(x - t);    a.push_back(x + t);    pq.push_back(a);}int getnum(){    int num = 0;    for(int i = 0; i < pq.size(); ++num)    {        float center = pq[i++][1];        while(i < pq.size())        {            if(pq[i][0] <= center)            {                if(pq[i][1] < center)                {                    center = pq[i][1];                }                i++;            }            else                break;        }    }    return num;}int main(){    scanf("%d%d", &n, &d);    while(n != 0 || d != 0)    {        casenum++;        pq.clear();        bool flag = true;        for(int i = 0; i < n; i++)        {            int x, y;            scanf("%d%d", &x, &y);            if(y <= d)                getradis(x, y);            else                flag = false;        }        if(flag)        {            sort(pq.begin(), pq.end());            printf("Case %d: %d\n", casenum, getnum());        }        else            printf("Case %d: %d\n", casenum, -1);        scanf("%d%d", &n, &d);    }    return 0;}