Periodic Signal

最新推荐文章于 2018-09-13 16:30:58 发布

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描述

Profess X is an expert in signal processing. He has a device which can send a particular 1 second signal repeatedly. The signal is A0 ... An-1 under n Hz sampling.

One day, the device fell on the ground accidentally. Profess X wanted to check whether the device can still work properly. So he ran another n Hz sampling to the fallen device and got B0 ... Bn-1.

To compare two periodic signals, Profess X define the DIFFERENCE of signal A and B as follow:

You may assume that two signals are the same if their DIFFERENCE is small enough.
Profess X is too busy to calculate this value. So the calculation is on you.

输入

The first line contains a single integer T, indicating the number of test cases.

In each test case, the first line contains an integer n. The second line contains n integers, A0 ... An-1. The third line contains n integers, B0 ... Bn-1.

T≤40 including several small test cases and no more than 4 large test cases.

For small test cases, 0<n≤6⋅103.

For large test cases, 0<n≤6⋅104.

For all test cases, 0≤Ai,Bi<220.

输出

For each test case, print the answer in a single line.

样例输入

2
9
3 0 1 4 1 5 9 2 6
5 3 5 8 9 7 9 3 2
5
1 2 3 4 5
2 3 4 5 1

样例输出

80
0





#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <climits>
#include <ctime>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread(ch) freopen(ch,"r",stdin)
#define fwrite(ch) freopen(ch,"w",stdout)


using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const double eps = 1e-3;
const int maxn = 61234;


struct Line
{
    LL l,r,val;
    bool operator <(const struct Line a)const
    {
        return l < a.l;
    }
} ln[maxn];


LL bit[maxn];
LL st,en,t;
int n,m,tp;
LL to[maxn];
map <LL,int> mp;
int id,iid;


int Lowbit(int x)
{
    return x&(-x);
}


void Add(int x,LL ad)
{
    x++;
    while(x <= id+1)
    {
        bit[x] += ad;
        x += Lowbit(x);
    }
}


LL Sum(int x)
{
    LL ans = 0;
    x++;
    while(x)
    {
        ans += bit[x];
        x -= Lowbit(x);
    }
    return ans;
}


int Search(LL x)
{
    int l,r;
    l = 0,r = id;
    int ans = -1;


    while(l <= r)
    {
        int mid = (l+r)>>1;
        if(to[mid] <= x)
        {
            ans = mid;
            l = mid+1;
        }
        else r = mid-1;
    }
    return ans;
}


LL solve()
{
    LL mx = 0;
    memset(bit,0,sizeof(bit));


    sort(ln,ln+tp);


    int l,r;
    l = 0,r = 0;


    for(int i = 0; i <= id; ++i)
    {
        int rr = Search(to[i]+t);
        while(r < tp && mp[ln[r].l] <= rr)
        {
            Add(mp[ln[r].r],ln[r].val);
            ++r;
        }


        while(l < tp && mp[ln[l].l] < i)
        {
            Add(mp[ln[l].r],-ln[l].val);
            ++l;
        }
        mx = max(mx,Sum(rr));
//        printf("%d,%d %lld\n",i,rr,mx);
//        printf("%d %d\n",l,r-1);
    }


    return mx;
}


int main()
{
    LL t2,x,y,val,sum;


    while(~scanf("%lld%lld",&t,&t2))
    {
        mp.clear();
        iid = 0;
        scanf("%lld",&st);
        en = st+t2;


        scanf("%d%d",&n,&m);


        tp = 0;
        sum = 0;
        for(int i = 0; i < n; ++i)
        {
            scanf("%lld%lld%lld",&x,&y,&val);
            x += y;
            if(st <= x && x <= en)
            {
                ln[tp].l = x+y;
                ln[tp].r = (en-x)/(2*y)*(2*y)+ln[tp].l;
                ln[tp++].val = val;
                sum += val;
                to[iid++] = ln[tp-1].l;
                to[iid++] = ln[tp-1].r;
//                printf("%lld %lld\n",ln[tp-1].l,ln[tp-1].r);
            }
        }


        for(int i = 0; i < m; ++i)
        {
            scanf("%lld%lld%lld",&x,&y,&val);


            ln[tp].l = x+y;


            if(st <= x+2*y && x+2*y <= en)
            {
                ln[tp].r = (en-x)/(2*y)*(2*y)+ln[tp].l;
            }
            else ln[tp].r = x+y;
            ln[tp++].val = val;
            sum += val;
            to[iid++] = ln[tp-1].l;
            to[iid++] = ln[tp-1].r;
       }


        sort(to,to+iid);


        id = -1;


        for(int i = 0; i < iid; ++i)
        {
            if(!i || to[i] != to[i-1])
            {
                ++id;
                to[id] = to[i];
                mp[to[i]] = id;
            }
        }


        printf("%lld\n",sum-solve());
    }




    return 0;
}