poj1328不完全解题报告

//
//  main.c
//  poj1328
//
//  Created by 韩雪滢 on 11/16/16.
//  Copyright © 2016 韩雪滢. All rights reserved.
/*
*不完全解题报告
*实现了解题思路，单个测试用例输入正确，没有写循环输入
*如发现输入结果错误，请在下面留言，不胜感激
*
*解题思路：
*贪心算法
*在读入每个岛屿的坐标时，同时计算以该岛屿为圆心，雷达范围半径为半径的圆与x轴相交的横坐标范围
*如果范围不重叠，雷达＋1
*/

#include <stdio.h>
#include <mm_malloc.h>
#include <math.h>

struct island{
    int x;
    int y;
    int start;
    int terminal;
};

struct radar{
    int x;
    int y;
    int radius;
};

struct island *inIslands;
struct radar *setRadars;
int *prior;


int radarNum(int n){
    int k,i,j,m;
    int result=1;
    
    for(k=0;k<n;k++)
        prior[k]=k;
    
    for(i=0;i<n;i++){
        for(j=i-1;(j>=0) && (inIslands[i].start < inIslands[prior[j]].start);j--)
        {
            prior[j+1]=prior[j];
        }
        prior[j+1]=i;
    }
    
    for(m=0;m<n-1;m++){
        if(inIslands[m].terminal<inIslands[m+1].start)
            result++;
    }
    return result;
}


int main(int argc, const char * argv[]) {
    int n,d;
    scanf("%d%d",&n,&d);
    inIslands = (struct island*)malloc(sizeof(struct island)*n);
    prior = (int*)malloc(sizeof(int)*n);
    int i;
    for(i=0;i<n;i++){
        scanf("%d%d",&inIslands[i].x,&inIslands[i].y);
        int temp = sqrt(d*d-inIslands[i].y*inIslands[i].y);
        inIslands[i].start = inIslands[i].x-temp;
        inIslands[i].terminal = inIslands[i].x+temp;
    }
    
    int result = radarNum(n);
    
    printf("结果:%d\n",result);
    
    int j;
    for(j=0;j<n;j++){
        printf("岛屿%d:［%d，%d]\n",prior[j],inIslands[prior[j]].start,inIslands[prior[j]].terminal);
    }
    
    return 0;
}