The beautiful values of the palace (扫描线）

最新推荐文章于 2020-07-10 22:50:39 发布

原创最新推荐文章于 2020-07-10 22:50:39 发布 · 183 阅读

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本文介绍了一种解决矩阵内点权值和问题的算法，采用扫描线技术进行高效计算。通过预处理和树状数组，实现了对矩阵范围内点权值的快速求和，适用于动态更新和查询场景。

题目

https://nanti.jisuanke.com/t/41298

题意

问矩阵内点的权值和

思路

扫描线

代码

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e6+100;
typedef long long ll;
ll num[maxn];
ll numx[maxn];
int ans[maxn];
struct point
{
    int x,y,w;
}po[maxn];
struct node
{
    int x1,x2,y,id,f;
}ju[maxn];
bool cmppo(point a,point b)
{
    return a.y < b.y;
}
bool cmpju(node a,node b)
{
    if(a.y == b.y)
    {
        return a.f < b.f;
    }
    return a.y < b.y;
}
int get(ll x)
{
    int ans = 0;
    while(x)
    {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}
int ok(int x,int y,int n)
{
    int a = n - x + 1;
    int b = y;
    int c = x;
    int d = n - y + 1;
    if(a <= b&&a <= c&&a <= d)
    {
        ll ans = num[a-1];
        ans += n - (y-1) - (a-1);
        return get(ans);
    }
    else if(b <= c&&b <= d)
    {
        ll ans = num[b-1] + numx[b];
        ans += n - (b-1) - (x-1);
        return get(ans);
    }
    else if(c <= d)
    {
        ll ans = num[c-1] + 2*numx[c];
        ans += y - (c - 1);
        return get(ans);
    }
    else
    {
        ll ans = num[d-1] + 3*numx[d];
        ans += x - (d-1);
        return get(ans);
    }
}


int tree[maxn*4];
void build(int l,int r,int x)
{
    tree[x] = 0;
    if(l==r) return ;
    int mid = (l+r)/2;
    build(l,mid,x*2);
    build(mid+1,r,x*2+1);
}
void updata(int x,int l,int r,int id,int val)
{
    if(l==r)
    {
        tree[x] += val;
        return ;
    }
    int mid = (l + r) / 2;
    if(id <= mid) updata(x*2,l,mid,id,val);
    else updata(x*2+1,mid+1,r,id,val);
    tree[x] = tree[x*2] + tree[x*2+1];
}
int query(int x,int l,int r,int L,int R)
{
    if(l==L&&r==R)
    {
        return tree[x];
    }
    int mid = (l+r)/2;
    if(R <= mid) return query(x*2,l,mid,L,R);
    else if(L > mid) return query(x*2+1,mid+1,r,L,R);
    else return query(x*2,l,mid,L,mid) + query(x*2+1,mid+1,r,mid+1,R);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m,q;
        scanf("%d%d%d",&n,&m,&q);
        int mm = (n+1)/2;
        num[0] = 0;
        int top = n;
        for(int i = 1;i <= mm;i++)
        {
            num[i] = num[i-1] + 8*(mm-i);
            numx[i] = n - 1 - 2*(i-1);
        }
        num[mm+1] = num[mm] + 1;
        for(int i = 1;i <= m;i++)
        {
            scanf("%d%d",&po[i].x,&po[i].y);
            po[i].w = ok(po[i].x,po[i].y,n);
        }
        for(int i = 1;i <= q;i++)
        {
            ans[i] = 0;
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            y1--;
            ju[i].x1 = x1,ju[i].x2 = x2,ju[i].y = y1,ju[i].f = -1;
            ju[i+q].x1 = x1,ju[i+q].x2 = x2,ju[i+q].y = y2,ju[i+q].f = 1;
            ju[i].id = ju[i+q].id = i;
        }
        sort(po+1,po+1+m,cmppo);
        sort(ju+1,ju+1+2*q,cmpju);
        int p = 1;
        build(1,top,1);
        int yy = ju[1].y;
        for(int i = 1;i <= 2*q;i++)
        {
            while(1)
            {
                if(p > m) break;
                if(po[p].y > ju[i].y) break;
                updata(1,1,top,po[p].x,po[p].w);
                p++;
            }
            int l = ju[i].x1;
            int r = ju[i].x2;
            ans[ju[i].id] += ju[i].f * query(1,1,top,l,r);
        }
        for(int i = 1;i <= q;i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}