把询问搞成4个，cdq分治。
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a;i <= b; i++) 3 #define drep(i, a, b) for (int i = a; i >= b; i--) 4 #define REP(i, a, b) for (int i = a; i < b; i++) 5 #define mp make_pair 6 #define pb push_back 7 #define clr(x) memset(x, 0, sizeof(x)) 8 #define xx first 9 #define yy second 10 using namespace std; 11 typedef long long i64; 12 typedef pair<int, int> pii; 13 const int inf = ~0U >> 1; 14 const i64 INF = ~0ULL >> 1; 15 //********************************** 16 17 const int maxn = 200005; 18 19 int c[2000005], w; 20 struct Complex { 21 int flag; 22 int x, y, c; 23 int id, ans; 24 int pos, l; 25 inline bool operator < (const Complex &a) const { 26 return x < a.x || 27 x == a.x && y < a.y || 28 x == a.x && y == a.y && c < a.c; 29 } 30 } src[maxn], t[maxn]; 31 32 inline void add(int x, int v) { 33 while (x <= w) { 34 c[x] += v; 35 x += x & -x; 36 } 37 } 38 inline int get(int x) { 39 int ret(0); 40 while (x > 0) { 41 ret += c[x]; 42 x -= x & -x; 43 } 44 return ret; 45 } 46 int ans[10005]; 47 48 void cdq(int l, int r) { 49 if (l == r) return; 50 int mid = l + r >> 1, l1 = l, l2 = mid + 1; 51 rep(i, l, r) { 52 if (src[i].id <= mid && !src[i].l) add(src[i].y, src[i].c); 53 if (src[i].id > mid && src[i].l) ans[src[i].pos] += src[i].l * get(src[i].y); 54 } 55 rep(i, l, r) if (src[i].id <= mid && !src[i].l) add(src[i].y, -src[i].c); 56 rep(i, l, r) if (src[i].id <= mid) t[l1++] = src[i]; else t[l2++] = src[i]; 57 memcpy(src + l, t + l, (r - l + 1) * sizeof(Complex)); 58 cdq(l, mid); cdq(mid + 1, r); 59 } 60 61 int main() { 62 int cnt(0), n(0), s; 63 scanf("%d%d", &s, &w); 64 int flag; 65 while (scanf("%d", &flag), flag ^ 3) { 66 if (flag == 1) { 67 ++n; 68 src[n].id = n; src[n].l = 0; src[n].pos = 0; 69 scanf("%d%d%d", &src[n].x, &src[n].y, &src[n].c); 70 } 71 else { 72 int x, y, a, b; scanf("%d%d%d%d", &x, &y, &a, &b); 73 74 ans[++cnt] = s * (a - x) * (b - y); 75 76 ++n; 77 src[n].id = n; src[n].l = 1; src[n].pos = cnt; 78 src[n].x = a, src[n].y = b, src[n].c = inf; 79 80 ++n; 81 src[n].id = n; src[n].l = -1; src[n].pos = cnt; 82 src[n].x = a, src[n].y = y - 1, src[n].c = inf; 83 84 ++n; 85 src[n].id = n; src[n].l = -1; src[n].pos = cnt; 86 src[n].x = x - 1, src[n].y = b, src[n].c = inf; 87 88 ++n; 89 src[n].id = n; src[n].l = 1; src[n].pos = cnt; 90 src[n].x = x - 1, src[n].y = y - 1, src[n].c = inf; 91 } 92 } 93 sort(src + 1, src + n + 1); 94 cdq(1, n); 95 rep(i, 1, cnt) printf("%d ", ans[i]); 96 return 0; 97 }