标签:hdu bestcoder round 枚举
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4932
3 3 1 2 3 3 1 2 4 4 1 9 100 10
1.000 2.000 8.000HintFor the first sample , a legal answer is [1,2] [2,3] so the length is 1. For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2. For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.
题意:
求最大能够覆盖所有所给的点的区间长度(所给的点必须处于区间两端)。
思路:
答案一定是相邻点之间的差值或者是相邻点之间的差值除以2,那么把这些可能的答案先算出来,然后依次从最大的开始枚举进行验证即可。
代码如下:
#include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int MAXN = 147; int f[MAXN];//记录线段方向 double p[MAXN]; double d[MAXN];//相邻断点的差值 int n; void init() { memset(p,0,sizeof(p)); memset(f,0,sizeof(f)); memset(d,0,sizeof(d)); } bool Judge(double tt) { int i; for(i = 1; i < n-1; i++) { if(p[i] - tt < p[i-1] && p[i] + tt > p[i+1]) break;//无论向左还是向右均为不符合 if(p[i] - tt >= p[i-1])//向左察看 { if(f[i-1] == 2)//如果前一个是向右的 { if(p[i] - p[i-1] == tt) f[i] = 1;//两个点作为线段的两个端点 else if(p[i] - p[i-1] >= 2*tt)//一个向左一个向右 { f[i] = 1; } else if(p[i] + tt <= p[i+1]) { f[i] = 2;//只能向右 } else return false; } else f[i] = 1; } else if(p[i] + tt <= p[i+1]) f[i] = 2; } if(i == n-1)//全部符合 return true; return false; } int main() { int t; scanf("%d",&t); while(t--) { init(); scanf("%d",&n); for(int i = 0; i < n; i++) { scanf("%lf",&p[i]); } sort(p,p+n); int cont = 0; for(int i = 1; i < n; i++) { d[cont++] = p[i] - p[i-1]; d[cont++] = (p[i] - p[i-1])/2.0; } sort(d,d+cont); double ans = 0; for(int i = cont-1; i >= 0; i--) { memset(f,0,sizeof(f)); f[0] = 1; //开始肯定是让线段向左 if(Judge(d[i])) { ans = d[i]; break; } } printf("%.3lf\n",ans); } return 0; }
hdu4932 Miaomiao's Geometry (BestCoder Round #4 枚举),布布扣,bubuko.com
hdu4932 Miaomiao's Geometry (BestCoder Round #4 枚举)
标签:hdu bestcoder round 枚举
原文地址:http://blog.csdn.net/u012860063/article/details/38496873