Bzoj 1176 Balkan Mokia CDQ Division

Some operations, give a coordinate to represent the point plus a number, and find out a rectangle of all the number of the and.

Idea: one-eye question, two-bit tree-shaped array of water over.

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What's wrong? w<=2000000. Tease me? Can this n^2 open?

At this time CDQ the god ox to help us again.

This problem should be regarded as the cdq of the template problem, simple analysis, in fact, not difficult.

Before writing this question, it is suggested to write the problem of Bzoj 1935 SHOI Tree gardener's troubles, which is a simplified version of the subject.

According to the normal solution, we should establish a two-bit data structure, and then maintain two-dimensional information separately. If you use a dynamically open line segment tree, you may get stuck, but you should know that you have written a two-dimensional segment tree. This is not as good as a two-bit tree array.

So we try to shrink one dimension and make it a one-dimensional problem. Consider both the weighted value and the query as an action, a query action is split into 4, and then the operations are sorted by x-coordinate. In the process of CDQ division, we guarantee that the first half of the last part is less than the second half of the time when X is ordered and the answer is updated. Then , after each processing [l,mid], use the weighted value action in [L,mid] to update the Ask action in [Mid + 1,r].

CODE:

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 2000010#define INF 0x3f3f3f3fusing namespace std;    struct complex{int flag;    int x,y,c;         int Id,ans;    BOOL operator < (const Complex &a) Const {return x < a.x; }}src[max],t[max]; int asks,cnt,size; int Fenwick[max],t,v[max];        inline void Fix (int x,int c) {for (; x <= size; x + = X&-x) {if (v[x]! = T) v[x] = t,fenwick[x] = 0;    FENWICK[X] + = C;    }} inline int getsum (int x) {int re = 0;    for (; x; x = x&-x) if (v[x] = = T) Re + = fenwick[x]; return re;}    void CDQ (int l,int r) {if (L = = r) return;    int mid = (L + r) >> 1;    int L1 = L,L2 = mid + 1;        for (int i = l; I <= R; ++i) if (src[i].id <= mid) t[l1++] = Src[i];    else t[l2++] = src[i];    memcpy (src + l,t + l,sizeof (Complex) * (r-l + 1));    CDQ (L,mid);    ++t;    int now = l; for (int i = mid +1; I <= R;                ++i) {while (src[now].x <= src[i].x && now <= mid) {if (Src[now].flag = = 1)            Fix (SRC[NOW].Y,SRC[NOW].C);        ++now;    } if (Src[i].flag = = 2) Src[i].ans + = Getsum (SRC[I].Y);    } CDQ (mid + 1,r);    L1 = L,l2 = mid + 1; for (int i = l; I <= R; ++i) if ((Src[l1] < SRC[L2] && L1 <= mid) | | L2 > R) t[i] = SR        C[l1++];    else t[i] = src[l2++]; memcpy (src + l,t + l,sizeof (Complex) * (r-l + 1));}    BOOL CMP (const Complex &a,const Complex &b) {return a.id < b.id;} int main () {scanf ("%*d%d", &size);    int flag,x1,y1,x2,y2;            while (scanf ("%d", &flag), flag^3) {if (flag = = 1) {Src[++cnt].flag = 1,src[cnt].id = cnt;        scanf ("%d%d%d", &src[cnt].x,&src[cnt].y,&src[cnt].c);            } else {scanf ("%d%d%d%d", &x1,&y1,&x2,&y2); Src[++cnt].flag = 2,src[cnt].id = Cnt                         src[cnt].x = X2,src[cnt].y = Y2,SRC[CNT].C = INF;            Src[++cnt].flag = 2,src[cnt].id = cnt;                         src[cnt].x = X1-1,src[cnt].y = Y2,SRC[CNT].C = INF;            Src[++cnt].flag = 2,src[cnt].id = cnt;                         src[cnt].x = X2,src[cnt].y = Y1-1,SRC[CNT].C = INF;            Src[++cnt].flag = 2,src[cnt].id = cnt;        src[cnt].x = X1-1,src[cnt].y = Y1-1,SRC[CNT].C = INF;    }} sort (src + 1,src + cnt + 1);    CDQ (1,CNT);    Sort (src + 1,src + cnt + 1,cmp);            for (int i = 1; I <= cnt; ++i) if (Src[i].flag = = 2) {int ans = 0;            Ans + = Src[i++].ans;            Ans-= Src[i++].ans;            Ans-= Src[i++].ans;            Ans + = Src[i].ans;        printf ("%d\n", ans); } return 0;}

Bzoj 1176 Balkan Mokia CDQ Division