CodeForces 610D Vika and Segments

题目链接:

http://codeforces.com/problemset/problem/610/D

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虽然说这题是线段并 但是如果会写矩形面积并的话就直接写矩形并不用考虑那么多了

不过这里额外说明下 写矩形面积并的线段树的时候

要注意到某一段 $ -1 $一定是在这一段$ +1 $之后才会出现的操作

因此标记就是这一段被“完全”覆盖的次数 并且部分修改时无需进行标记下放

这题作为矩形面积并的第一道题也是很适合的

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <algorithm>
 5 using namespace std;
 6 const int N = 1e5 + 10;
 7 struct rec
 8 {
 9     int xa, ya, xb, yb;
10 }a[N];
11 struct line
12 {
13     int x, ya, yb, num;
14 }b[N << 1];
15 int hash[N << 1];
16 int sum[N << 3], flag[N << 3];
17 int n;
18 long long ans = 0;
19 bool cmp (const line &aa, const line &bb)
20 {
21     return aa.x < bb.x;
22 }
23 void pushup(int x, int tl, int tr)
24 {
25     if(flag[x])
26         sum[x] = hash[tr + 1] - hash[tl];
27     else if(tl != tr)
28         sum[x] = sum[x << 1] + sum[x << 1 | 1];
29     else
30         sum[x] = 0;
31     return;
32 }
33 void update(int x, int L, int R, int tl, int tr, int num)
34 {
35     if(L <= tl && R >= tr)
36     {
37         flag[x] += num;
38         pushup(x, tl, tr);
39         return;
40     }
41     int mid = (tl + tr) >> 1;
42     if(L <= mid)
43         update(x << 1, L, R, tl, mid, num);
44     if(R > mid)
45         update(x << 1 | 1, L, R, mid + 1, tr, num);
46     pushup(x, tl, tr);
47 }
48 int main()
49 {
50     scanf("%d", &n);
51     for(int i = 1; i <= n; ++i)
52     {
53         scanf("%d%d%d%d", &a[i].xa, &a[i].ya, &a[i].xb, &a[i].yb);
54         if(a[i].xa > a[i].xb)
55             swap(a[i].xa, a[i].xb);
56         if(a[i].ya > a[i]. yb)
57             swap(a[i].ya, a[i].yb);
58         ++a[i].xb;
59         ++a[i].yb;
60         hash[i * 2 - 1] = a[i].ya;
61         hash[i * 2] = a[i].yb;
62         b[i * 2 - 1].x = a[i].xa;
63         b[i * 2 - 1].num = 1;
64         b[i * 2].x = a[i].xb;
65         b[i * 2].num = -1;
66         b[i * 2].ya = b[i * 2 - 1].ya = a[i].ya;
67         b[i * 2].yb = b[i * 2 - 1].yb = a[i].yb;
68     }
69     sort(hash + 1, hash + 1 + n * 2);
70     sort(b + 1, b + 1 + n * 2, cmp);
71     b[0].x = b[1].x;
72     for(int i = 1; i <= n * 2; ++i)
73     {
74         int L, R;
75         L = lower_bound(hash + 1, hash + 1 + n * 2, b[i].ya) - hash;
76         R = lower_bound(hash + 1, hash + 1 + n * 2, b[i].yb) - hash - 1;
77         ans += (long long) (b[i].x - b[i - 1].x) * sum[1];
78         update(1, L, R, 1, n * 2 - 1, b[i].num);
79     }
80     printf("%lld\n", ans);
81     return 0;
82 }