nyoj 287(区间覆盖)

来源：互联网发布：编程常用的算法编辑：程序博客网时间：2024/06/05 19:57

题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=287

解题思路：首先求出来每个点的临界区域，即这个圆心能够将其覆盖的范围。。。求出了每个点的覆盖区域，那么问题就转化为区间的覆盖问题了。。。在算重叠的部分花了好长的时间而且还没有写好，还是没有把出现的情况讨论清楚。。。

AC:

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;struct node{double left,right;}a[1005];bool cmp(node a,node b){return a.left < b.left;}int main(){int n,d,cas = 1;bool flag;while(scanf("%d%d",&n,&d)!=EOF && n + d){flag = false;for(int i = 0; i < n; i++){int x,y;scanf("%d%d",&x,&y);double tmp = double(d*d - y*y);if(tmp >= 0){tmp = sqrt(tmp);a[i].left = x - tmp;a[i].right = x + tmp;}else flag = true;}if(flag){printf("Case %d: -1\n",cas++);continue;}sort(a,a+n,cmp);int cnt = 1;double t = a[0].right;for(int i = 1; i < n; i++){if(a[i].left > t){cnt++;t = a[i].right;}else{if(a[i].right < t)t = a[i].right;}}printf("Case %d: %d\n",cas++,cnt);}return 0;}

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