Day6 - L - Mokia HYSBZ - 1176

维护一个W*W的矩阵，初始值均为S.每次操作可以增加某格子的权值,或询问某子矩阵的总权值.修改操作数M<=160000,询问数Q<=10000,W<=2000000.

Input

第一行两个整数,S,W;其中S为矩阵初始值;W为矩阵大小

接下来每行为一下三种输入之一(不包含引号):

"1 x y a"

"2 x1 y1 x2 y2"

"3"

输入1:你需要把(x,y)(第x行第y列)的格子权值增加a

输入2:你需要求出以左下角为(x1,y1),右上角为(x2,y2)的矩阵内所有格子的权值和,并输出

输入3:表示输入结束

Output

对于每个输入2,输出一行,即输入2的答案

Sample Input0 4 1 2 3 3 2 1 1 3 3 1 2 2 2 2 2 2 3 4 3

Sample Output3 5Hint

 

保证答案不会超过int范围

思路：三元偏序问题，CDQ分治，(time,x,y)，将每一次查询拆成四个点即可，初始值预处理一下就行

using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 2e6+10;

int C[maxm], S, W, ans[maxm];

void add(int x, int val) {
    for(; x <= maxm; x += lowbit(x))
        C[x] += val;
}

int getsum(int x) {
    int ret = 0;
    for(; x; x -= lowbit(x))
        ret += C[x];
    return ret;
}

void clearr(int x) {
    for(; x <= maxm; x += lowbit(x)) {
        if(C[x] == 0) break;
        C[x] = 0;
    }
}

struct Node {
    int type, tim, x, y, w, id;
} buf[maxm], res[maxm];

bool cmp_S(Node a, Node b) {
    if(a.tim == b.tim && a.x == b.x) return a.y < b.y;
    if(a.tim == b.tim) return a.x < b.x;
    return a.tim < b.tim;
}


void CDQ(int L, int R) {
    if(L >= R) return;
    int mid = L+R >> 1;
    CDQ(L, mid), CDQ(mid+1, R);
    int i = L, j = mid+1, k = L;
    //左修改右查询
    while(i <= mid && j <= R) {
        if(res[i].x <= res[j].x) {
            if(res[i].type == 1)
                add(res[i].y, res[i].w);
            buf[k++] = res[i++];
        } else {
            if(res[j].type == 2)
                ans[res[j].id] += getsum(res[j].y) * res[j].w;
            buf[k++] = res[j++];
        }
    }
    while(i <= mid)
        buf[k++] = res[i++];
    while(j <= R) {
        ans[res[j].id] += getsum(res[j].y) * res[j].w;
        buf[k++] = res[j++];
    }
    for(int t = L; t <= R; ++t) {
        clearr(buf[t].y); // 清空树状数组
        res[t] = buf[t];
    }
}

int main() {
    scanf("%d%d", &S, &W);
    int type, sz = 0, x1, x2, y1, y2, tim = 0, query = 0;
    while(scanf("%d", &type) && type != 3) {
        if(type == 1) {
            scanf("%d%d%d", &res[sz].x, &res[sz].y, &res[sz].w);
            res[sz].tim = tim++;
            res[sz++].type = type;
        } else if(type == 2) {
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            ans[++query] = (y2 - y1 + 1) * (x2 - x1 + 1) * S;
            res[sz++] = {type, tim++, x1-1, y1-1, 1, query};
            res[sz++] = {type, tim++, x2, y1-1, -1, query};
            res[sz++] = {type, tim++, x1-1, y2, -1, query};
            res[sz++] = {type, tim++, x2, y2, 1, query};
        }
    }
    sort(res, res+sz, cmp_S);
    CDQ(0, sz-1);
    for(int i = 1; i <= query; ++i)
        printf("%d\n", ans[i]);
    return 0;
}