题目
Cogs
没有Bzoj的权限号
Sol
离线，\(CDQ\)分治，把询问拆成\(4\)个，变成每次求二位前缀和
那么只要一个修改操作(关键字为时间，\(x\)，\(y\))都在这个询问前，就可以累计答案
那么就成了偏序问题了，直接\(CDQ\)
注意当\(x\)相等时要把修改丢在前面
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
using namespace std;
typedef long long ll;
const int _(2e6 + 5);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, cal[_], Q;
ll ans[_], bit[_], a[_];
struct Data{
int id, x, y, op;
} q[_], tmp[_];
IL void Add(RG int x, RG ll v){
for(; x <= n; x += x & -x) bit[x] += v;
}
IL ll Query(RG int x){
RG ll ret = 0;
for(; x; x -= x & -x) ret += bit[x];
return ret;
}
IL bool Cmp(RG int x, RG int y){
if(q[x].x != q[y].x) return q[x].x < q[y].x;
if(!q[x].op) return 1;
return 0;
}
IL void CDQ(RG int l, RG int r){
if(l == r) return;
RG int mid = (l + r) >> 1;
CDQ(l, mid); CDQ(mid + 1, r);
for(RG int i = l, j = mid + 1, k = l; k <= r; ++k)
if(j > r || (i <= mid && Cmp(i, j))) tmp[k] = q[i++];
else tmp[k] = q[j++];
for(RG int i = l; i <= r; ++i) q[i] = tmp[i];
for(RG int i = l; i <= r; ++i)
if(q[i].id <= mid){
if(!q[i].op) Add(q[i].y, a[q[i].id]);
}
else{
if(q[i].op){
ans[cal[q[i].id]] += 1LL * Query(q[i].y) * q[i].op;
}
}
for(RG int i = l; i <= r; ++i)
if(q[i].id <= mid && !q[i].op) Add(q[i].y, -a[q[i].id]);
}
int main(RG int argc, RG char* argv[]){
File("mokia");
Input(); n = Input();
for(RG int op = Input(); op != 3; op = Input()){
if(op == 1){
RG int x = Input(), y = Input(); a[++m] = Input();
q[m] = (Data){m, x, y, 0};
}
else{
RG int x1 = Input(), y1 = Input(), x2 = Input(), y2 = Input();
++Q;
q[++m] = (Data){m, x2, y2, 1}; cal[m] = Q;
q[++m] = (Data){m, x1 - 1, y1 - 1, 1}; cal[m] = Q;
q[++m] = (Data){m, x1 - 1, y2, -1}; cal[m] = Q;
q[++m] = (Data){m, x2, y1 - 1, -1}; cal[m] = Q;
}
}
CDQ(1, m);
for(RG int i = 1; i <= Q; ++i) printf("%lld\n", ans[i]);
return 0;
}