hdu 5920 Wool 思路

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≤105,1≤L≤R≤1018).

The second line contains n integers, the i-th integer denotes ai (1≤ai≤1018).

 

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

Hint

In the first example, $ 2, 3 $ are available. In the second example, $ 6, 7, 8, 9, 10 $ are available.

思路：求出每条边的合法区间，区间合并一下，根据L，R，求ans；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=1e5+10,M=1e7+10,inf=1e9+10;
const ll mod=998244353;
ll a[N];
struct is
{
    ll x,y;
}gg[N];
int cmp(is x,is y)
{
    if(x.y!=y.y)
    return x.y<y.y;
    return x.x<y.x;
}
ll check(ll x,ll y,ll l,ll r)
{
    ll maxx=max(x,l);
    ll minn=min(r,y);
    if(maxx<=minn)
    return minn-maxx+1;
    return 0;
}
is he(ll x,ll y,ll l,ll r)
{
    is ans;
    ans.x=min(x,l);
    ans.y=max(r,y);
    return ans;
}
int main()
{
    ll x,y,z,i,t;
    int T;
    ll L,R;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d",&x,&L,&R);
        for(i=1;i<=x;i++)
        scanf("%I64d",&a[i]);
        sort(a+1,a+x+1);
        for(i=1;i<x;i++)
        {
            gg[i].x=a[i+1]-a[i]+1;
            gg[i].y=a[i+1]+a[i]-1;
        }
        sort(gg+1,gg+x,cmp);
        int ji=0;
        gg[ji].x=gg[1].x;
        gg[ji].y=gg[1].y;
        ji++;
        for(i=2;i<x;i++)
        {
            if(gg[i].x<=gg[ji-1].y)
            {
                gg[ji-1]=he(gg[i].x,gg[i].y,gg[ji-1].x,gg[ji-1].y);
            }
            else
            {
                gg[ji].x=gg[i].x;
                gg[ji].y=gg[i].y;
                ji++;
            }
        }
        ll ans=0;
        for(i=0;i<ji;i++)
        {
            ans+=check(gg[i].x,gg[i].y,L,R);
        }
        printf("%I64d
",R-L+1-ans);
    }
    return 0;
}