hdu-5720 Wool(区间并+扫描线)

题目链接：

Wool

Time Limit: 8000/4000 MS (Java/Others)   

 Memory Limit: 262144/262144 K (Java/Others)

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 T (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≤10^5,1≤L≤R≤10^18).

The second line contains n integers, the i-th integer denotes ai (1≤ai≤10^18).

 

Output

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

 

Sample Input

2

2 1 3

1 1

4 3 10

1 1 2 4

 

Sample Output

2

5

 

题意：

 

已经由n跟长为ai的线段,现在给定区间[l,r],问这个区间里还有多少个数与那一堆数不会形成三角形;

 

思路：

 

把a数组排序,i<j;选a[i]和a[j],(a[i]-a[j],a[i]+a[j]);这是不能出现的数的区间,可知(a[i]-a[i-1],a[i]+a[i-1])是选a[i]这个数会出现的最大范围的区间;

所以就形成了n-1个区间[a[i]-a[i-1]+1,a[i]+a[i-1]-1],把这些区间扫一遍和并后求出覆盖了[l,r]中的多少个点,然后用[l,r]的点数减去就是答案了;

 

AC代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=998244353;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=1e3+10;
const double eps=1e-6;

int n;
LL l,r,a[N];
struct node
{
    LL l,r;
}po[N];
int cmp(node x,node y)
{
    if(x.l==y.l)return x.r<y.r;
    return x.l<y.l;
}
int vis[N];
int main()
{
        int t;
        read(t);
        while(t--)
        {
            mst(vis,0);
            read(n);read(l);read(r);
            For(i,0,n-1)read(a[i]);
            sort(a,a+n);
            For(i,1,n-1)
            {
                po[i].r=a[i]+a[i-1]-1;
                po[i].l=a[i]-a[i-1]+1;
            }
            sort(po+1,po+n,cmp);
            LL len=0;
            For(i,1,n-1)
            {
                if(i==n-1)continue;
                if(po[i].r<po[i+1].l)continue;
                else 
                {
                    po[i+1].l=po[i].l;
                    po[i+1].r=max(po[i].r,po[i+1].r);
                    vis[i]=1;
                }
            }
            For(i,1,n)
            {
                if(!vis[i])
                {
                    if(po[i].l>r||po[i].r<l||po[i].l>po[i].r)continue;
                    LL L=max(po[i].l,l),R=min(po[i].r,r);
                    len=len+(R-L+1);
                }
            }
            cout<<r-l+1-len<<"
";
        }
        
        return 0;
}