POJ2155 Matrix 二维线段树

关键词：线段树

二维线段树维护一个 维护一个X线段的线段树，每个X节点维护一个 维护一个Y线段的线段树。

注意，以下代码没有PushDownX。因为如果要这么做，PushDownX时，由于当前X节点的子节点可能存在标记，而标记不能叠加，导致每次PushDownX时都要把子节点PushDownX一次。每次PushDownX都要对子节点UpdateY，代价太高。这种情况下原则是：只有查询操作<<更新操作时才PushDownX。

#include <cstdio>
#include <cstring>
#include <cassert>
#include <vector>
using namespace std;

const int MAX_X = 2000, MAX_Y = 2000;

struct RangeTree2d
{
private:
    bool Rev[MAX_X * 4][MAX_Y * 4];
    int YlMin, YrMax, XlMin, XrMax;

    void PushDownY(int xCur, int yCur)
    {
        if (Rev[xCur][yCur])
        {
            Rev[xCur][yCur * 2] ^= 1;
            Rev[xCur][yCur * 2 + 1] ^= 1;
            Rev[xCur][yCur] = false;
        }
    }

    void UpdateY(int xCur, int yCur, int sl, int sr, int al, int ar)
    {
        assert(sl <= sr&&al <= ar&&al <= sr&&ar >= sl);
        if (al <= sl&&sr <= ar)
        {
            Rev[xCur][yCur] ^= 1;
            return;
        }
        int mid = (sr - sl) / 2 + sl;
        if (al <= mid)
            UpdateY(xCur, yCur * 2, sl, mid, al, ar);
        if (ar > mid)
            UpdateY(xCur, yCur * 2 + 1, mid + 1, sr, al, ar);
    }

    void UpdateY(int xCur, int l, int r)
    {
        UpdateY(xCur, 1, YlMin, YrMax, l, r);
    }

    bool QueryY(int xCur, int yCur, int l, int r, int y)
    {
        if (l == r)
            return Rev[xCur][yCur];
        PushDownY(xCur, yCur);
        int mid = (l + r) / 2;
        if (y <= mid)
            return QueryY(xCur, yCur * 2, l, mid, y);
        else
            return QueryY(xCur, yCur * 2 + 1, mid + 1, r, y);
    }

    bool QueryY(int xCur, int y)
    {
        return QueryY(xCur, 1, YlMin, YrMax, y);
    }

    void UpdateX(int xCur, int sl, int sr, int al, int ar, int yl, int yr)
    {
        //printf("C x:sl %d sr %d al %d  ar %d
", sl, sr, al, ar);
        assert(sl <= sr&&al <= ar&&al <= sr&&ar >= sl);
        if (al <= sl&&sr <= ar)
        {
            UpdateY(xCur, yl, yr);
            return;
        }
        int mid = (sr - sl) / 2 + sl;
        if (al <= mid)
            UpdateX(xCur * 2, sl, mid, al, ar, yl, yr);
        if (ar > mid)
            UpdateX(xCur * 2 + 1, mid + 1, sr, al, ar, yl, yr);
    }

    void QueryX(int xCur, int l, int r, int x, int y, bool &ans)
    {
        ans ^= QueryY(xCur, y);
        if (l == r)
            return;
        int mid = (l + r) / 2;
        if (x <= mid)
            QueryX(xCur * 2, l, mid, x, y, ans);
        else
            QueryX(xCur * 2 + 1, mid + 1, r, x, y, ans);
    }

public:
    void Init(int xlMin, int xrMax, int ylMin, int yrMax)
    {
        XlMin = xlMin;
        XrMax = xrMax;
        YlMin = ylMin;
        YrMax = yrMax;
        memset(Rev, false, sizeof(Rev));
    }

    void Update(int xl, int xr, int yl, int yr)
    {
        UpdateX(1, XlMin, XrMax, xl, xr, yl, yr);
    }

    int Query(int x, int y)
    {
        bool ans = false;
        QueryX(1, XlMin, XrMax, x, y, ans);
        return ans;
    }
}g;

int main()
{
#ifdef _DEBUG
    freopen("c:\noi\source\input.txt", "r", stdin);
#endif
    int totCase;
    scanf("%d", &totCase);
    while (totCase--)
    {
        int mSize, qCnt;
        scanf("%d%d", &mSize, &qCnt);
        g.Init(1, mSize, 1, mSize);
        while (qCnt--)
        {
            char op;
            int x1, x2, y1, y2;
            scanf("
%c", &op);
            switch (op)
            {
            case 'C':
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                g.Update(x1, x2, y1, y2);
                break;
            case 'Q':
                scanf("%d%d", &x1, &y1);
                printf("%d
", g.Query(x1, y1));
                break;
            default:
                assert(0);
            }
        }
        printf("
");
    }
    return 0;
}

反思：

1.不用动态开点，因为内存够。动态申请内存费时间。

2.不要将问题扩大化为求区域面积，提高了编程复杂度。