矩形面积交 [区间树状数组, 扫描线]

$矩形面积交$

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$color{red}{正解部分}$

每个矩形提出其 左边 和 右边 , 每条边都使用三元组 $(l, r, flag, type)$表示, 放置对应的 $x$ 坐标上,

使用一条 扫描线 从左向右扫, 如下图计算 橙色询问矩形 的答案,

计算相交面积 $s_2$ 时, 考虑使用 $y$ 总相交面积 $s_3$ 减去 $y$ 相交 $x$ 不相交面积 $s_1$ .

$s_3$ 可以在 询问矩形 右边 通过查询 $[l, r]$ 的面积和得到 .

使用前面 黄色虚线面积 减去 绿色面积 得到 $s_1$ .

区间树状数组 :

• 区间查询: $[1, r]$, $res = sumlimits_{i=1}^{r}sumlimits_{j=1}^i d[j] = sumlimits_{i=1}^{r} (r-i+1) imes d[i] = (r+1)sumlimits_{i=1}^{r} d[i] - sumlimits_{i=1}^{r}i imes d[i]$ .

  于是 维护 $d[i]$, $i imes d[i]$

• 区间修改 $(+x)$:
  对 $d[i]$ 差分, $l$ 处 $+x$, $r+1$ 处 $-x$ .
  对 $i imes d[i]$ 差分, $l$ 处 $+ x imes l$, $r+1$ 处 $- x imes (r+1)$ .

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$color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register
typedef long long ll;

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 5e5 + 5;

int N;
int M;
int W;
int L;
int cnt;

ll Ans[maxn];

struct Sgment{ int x, l, r, flag, id; } A[maxn<<2];

bool cmp(Sgment a, Sgment b){ return a.x < b.x; }

struct Bit_Tree{

        int lim;
        ll d[maxn], d2[maxn];

        void Add(int k, ll x){
                for(reg int i = k; i <= lim; i += (i&-i)) d[i] += x, d2[i] += 1ll*x*k;
        }

        ll Ask(int k){
                if(k <= 0) return 0;
                ll s = 0;
                for(reg int i = k; i; i -= (i&-i)) s += (1ll*k+1)*d[i] - d2[i];
                return s;
        }

        void modify(const int &l, const int &r, const ll &x){ Add(l, x), Add(r+1, -x); }

        ll query(const int &l, const int &r){ return Ask(r) - Ask(l-1); }

} bit_t[2];

int main(){
        N = read(), M = read(), W = read(), L = read();
        for(reg int i = 1; i <= N; i ++){
                int x1 = read(), y1 = read(), x2 = read(), y2 = read();
                A[++ cnt] = (Sgment){ x1, y1+1, y2, 1, 0 };
                A[++ cnt] = (Sgment){ x2, y1+1, y2, -1, 0 };
        }
        for(reg int i = 1; i <= M; i ++){
                int x1 = read(), y1 = read(), x2 = read(), y2 = read();
                A[++ cnt] = (Sgment){ x1, y1+1, y2, -1, i };
                A[++ cnt] = (Sgment){ x2, y1+1, y2, 1, i };
        }
        bit_t[0].lim = bit_t[1].lim = L;
        std::sort(A+1, A+cnt+1, cmp);
        for(reg int i = 0, j = 1; i <= W; i ++){
                while(A[j].x == i){
                        if(!A[j].id){
                                bit_t[1].modify(A[j].l, A[j].r, 1ll*A[j].flag*i);
                                bit_t[0].modify(A[j].l, A[j].r, A[j].flag);
                        }else{
                                ll s1 = bit_t[1].query(A[j].l, A[j].r);
                                ll s2 = bit_t[0].query(A[j].l, A[j].r);
                                Ans[A[j].id] += 1ll*A[j].flag*(1ll*i*s2 - s1);
                        }
                        j ++;
                }
        }
        for(reg int i = 1; i <= M; i ++) printf("%lld
", Ans[i]);
        return 0;
}