HDU5270Wool (区间扫描)

最新推荐文章于 2017-10-18 21:23:57 发布

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本文介绍了一个算法问题，任务是计算在给定长度范围内的新木棒数量，这些木棒不能与现有木棒构成三角形。文章提供了问题描述、输入输出格式、示例以及一种高效的求解方法。

Wool

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 918 Accepted Submission(s): 270

Problem Description

At dawn, Venus sets a second task for Psyche.

She is to cross a river and fetch golden wool from violent sheep who graze on the other side.

The sheep are wild and tameless, so Psyche keeps on throwing sticks to keep them away.

There are

n sticks on the ground, the length of the

i-th stick is

ai.

If the new stick she throws forms a triangle with any two sticks on the ground, the sheep will be irritated and attack her.

Psyche wants to throw a new stick whose length is within the interval

[L,R]. Help her calculate the number of valid sticks she can throw next time.

Input

The first line of input contains an integer

(1≤T≤10), which denotes the number of test cases.

For each test case, the first line of input contains single integer

n,L,R

(2≤n≤105,1≤L≤R≤1018).

The second line contains

n integers, the

i-th integer denotes

(1≤ai≤1018).

Output

For each test case, print the number of ways to throw a stick.

Sample Input

Sample Output

2
5

Hint
In the first example, $ 2, 3 $ are available.

In the second example, $ 6, 7, 8, 9, 10 $ are available.

题意：

给定一个大区间[L,R]，以及n根木棒的长度，问大区间内有多少个数可以不与现有的木棒构成三角形。

题解：

我们考察任意的两根木棒x,y(x>y)，我们分析反面，可以构成三角形的情况，那么能让x,y,z构成三角形的条件是x+y>z且x-y<z，也就是说当z落在(x-y,x+y)区间内的时候，会构成三角形，那么对于某个x,在比他小的y里面我们找一个最大的y即可快速计算出一个区间为[x-y+1,x+y-1]，之后我们再在大区间里抹去这些小区间即可。

代码：

//************************************************************************//
//*Author : Handsome How                                                 *//
//************************************************************************//
#include <cstdio>
#include <algorithm>
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
using namespace std;
typedef long long ll;
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
const int maxn = 1e5+5;
ll len[maxn], ans;
void del(ll L, ll R, ll l, ll r){	//从[L,R]里面抹去[l,r] 
	if(l>R||r<L) return;
	l = max(l,L);
	r = min(r,R);
	ans = ans - (r-l+1);
}

int main(){
	int T;
	gn(T);
	while(T--){
		int n;
		ll L,R;
		gn(n);gn(L);gn(R);
		fur(i,1,n)gn(len[i]);
		sort(len+1,len+1+n);
		ans = R - L + 1ll;		
		ll l = len[2]-len[1]+1ll, r = len[2]+len[1]-1ll;
		int now = 3;
		for(;now<=n;now++){
			if(len[now]-len[now-1]+1ll>r){
				del(L,R,l,r);
				l = len[now]-len[now-1]+1ll;
			}
			else{
				l = min(l,len[now]-len[now-1]+1ll);
			}
			r = len[now]+len[now-1]-1ll;
		}
		del(L,R,l,r);
		printf("%I64d\n",ans);
	}
	return 0;
}