【小练】树状数组&线段树

做了$HDU$上几个，因为$ACM\ step$里遇到了。。。把之前没提交过的两个拣出来护一下。。。

$HDU1892\ \ See\ you~$
- 一个矩形，初始每个格子有一本书，然后有一些移动，删除，统计的操作。
- 二维树状数组统计一下即可。

/* **********************************************

    File Name: 1892.cpp

    Auther: zhengdongjian@tju.edu.cn

    Created Time: 2015/9/23 星期三 下午 7:41:15

*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 1024;
int sum[MAX][MAX];

inline int lb(int& x) {
    return x & (-x);
}

void add(int x, int y, int c) {
    ++x;
    ++y;
    for (int i = x; i < MAX; i += lb(i)) {
        for (int j = y; j < MAX; j += lb(j)) {
            sum[i][j] += c;
        }
    }
}

int query(int x, int y) {
    ++x;
    ++y;
    int s = 0;
    for (int i = x; i > 0; i -= lb(i)) {
        for (int j = y; j > 0; j -= lb(j)) {
            s += sum[i][j];
        }
    }
    return s;
}

int main() {
    int T, Q;
    scanf(" %d", &T);
    for (int cas = 1; cas <= T; ++cas) {
        printf("Case %d:\n", cas);
        scanf(" %d", &Q);
        memset(sum, 0, sizeof(sum));
        for (int i = 0; i <= 1000; ++i) {
            for (int j = 0; j <= 1000; ++j) {
                add(i, j, 1);
            }
        }
        char op;
        int x1, y1, x2, y2, n1;
        while (Q--) {
            scanf(" %c", &op);
            if (op == 'S') {
                scanf(" %d %d %d %d", &x1, &y1, &x2, &y2);
                if (x1 > x2) {
                    swap(x1, x2);
                }
                if (y1 > y2) {
                    swap(y1, y2);
                }
                printf("%d\n", query(x2, y2) - query(x1 - 1, y2) - query(x2, y1 - 1) + query(x1 - 1, y1 - 1));
            } else if (op == 'A') {
                scanf(" %d %d %d", &x1, &y1, &n1);
                add(x1, y1, n1);
            } else if (op == 'D') {
                scanf(" %d %d %d", &x1, &y1, &n1);
                int num = query(x1, y1) - query(x1 - 1, y1) - query(x1, y1 - 1) + query(x1 - 1, y1 - 1);
                if (n1 > num) n1 = num;
                add(x1, y1, -n1);
            } else {
                scanf(" %d %d %d %d %d", &x1, &y1, &x2, &y2, &n1);
                int num = query(x1, y1) - query(x1 - 1, y1) - query(x1, y1 - 1) + query(x1 - 1, y1 - 1);
                if (n1 > num) n1 = num;
                add(x1, y1, -n1);
                add(x2, y2, n1);
            }
        }
    }
    return 0;
}

$HDU1514\ Stars$
- 好多星星，每个星星有等级，定义为左下区域的星星数目。
- 考虑离线处理，按照星星$y$坐标优先从小到大，$y$坐标相同时$x$坐标从小到大依次添加星星，则每次添加一个星星时，其左下区域的所有星星一定都已经添加完，并且所有在其上方($y$坐标大于它)的点都还未加入，于是只需要统计当前所有已经加入的点中，有多少横坐标小于它即可。基于上述思考，便可以在一维树状数组中完成查询。

/* **********************************************

    File Name: 1541.cpp

    Auther: zhengdongjian@tju.edu.cn

    Created Time: 2015/9/23 星期三 下午 8:46:50

*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 32007;
int sum[MAX];
int level[MAX >> 1];
P a[MAX];

inline int lb(int& x) {
    return x & (-x);
}

int query(int x) {
    ++x;
    int res = 0;
    while (x > 0) {
        res += sum[x];
        x -= lb(x);
    }
    return res;
}

void update(int x, int c) {
    ++x;
    while (x < MAX) {
        sum[x] += c;
        x += lb(x);
    }
}

int main() {
    int n;
    while (~scanf(" %d", &n)) {
        for (int i = 0; i < n; ++i) {
            scanf(" %d %d", &a[i].first, &a[i].second);
            swap(a[i].first, a[i].second);
        }
        sort(a, a + n);

        memset(sum, 0, sizeof(sum));
        memset(level, 0, sizeof(level));
        for (int i = 0; i < n; ++i) {
            ++level[query(a[i].second)];
            update(a[i].second, 1);
        }
        for (int i = 0; i < n; ++i) {
            printf("%d\n", level[i]);
        }
    }
    return 0;
}

$HDU1166\ 敌兵布阵$
- 维护一个区间和即可

/* **********************************************
    File Name: 1166_2.cpp
    Auther: zhengdongjian@tju.edu.cn
    Created Time: 2015/9/23 星期三 下午 8:33:05
*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 50007;
int sum[MAX << 2], a[MAX];

void build(int root, int left, int right) {
    if (left == right) {
        sum[root] = a[left];
        return;
    }
    int mid = (left + right) >> 1;
    build(root << 1, left, mid);
    build(root << 1 | 1, mid + 1, right);
    sum[root] = sum[root << 1] + sum[root << 1 | 1];
}

void update(int root, int left, int right, int pos, int num) {
    if (pos < left || pos > right) {
        return;
    } else if (left == right) {
        if (left == pos) {
            sum[root] += num;
        }
        return;
    }
    int mid = (left + right) >> 1;
    update(root << 1, left, mid, pos, num);
    update(root << 1 | 1, mid + 1, right, pos, num);
    sum[root] = sum[root << 1] + sum[root << 1 | 1];
}

int query(int root, int left, int right, int from, int to) {
    if (from > right || to < left) {
        return 0;
    } else if (from <= left && to >= right) {
        return sum[root];
    }

    int mid = (left + right) >> 1;
    return query(root << 1, left, mid, from, to) + query(root << 1 | 1, mid + 1, right, from, to);
}

int main() {
    int T;
    scanf(" %d", &T);
    for (int cas = 1; cas <= T; ++cas) {
        printf("Case %d:\n", cas);
        int n;
        scanf(" %d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf(" %d", a + i);
        }
        build(1, 1, n);
        char op[16];
        int le, ri, nu;
        while (~scanf(" %s", op)) {
            if (op[0] == 'E') {
                break;
            } else if (op[0] == 'A') {
                scanf(" %d %d", &le, &nu);
                update(1, 1, n, le, nu);
            } else if (op[0] == 'S') {
                scanf(" %d %d", &le, &nu);
                update(1, 1, n, le, -nu);
            } else {
                scanf(" %d %d", &le, &ri);
                printf("%d\n", query(1, 1, n, le, ri));
            }
        }
    }
    return 0;
}

$HDU3584\ Cube$
- 显然只需要维护每个点$not$操作的次数即可，若为偶数次则为$0$，否则为$1$。三维树状数组维护即可。

/* **********************************************

    File Name: 3584.cpp

    Auther: zhengdongjian@tju.edu.cn

    Created Time: 2015/9/23 星期三 下午 8:58:44

*********************************************** */
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 107;
int sum[MAX][MAX][MAX];

inline int lb(int& x) {
    return x & (-x);
}

int query(int x, int y, int z) {
    //++x, ++y, ++z;
    int s = 0;
    for (int i = x; i > 0; i -= lb(i)) {
        for (int j = y; j > 0; j -= lb(j)) {
            for (int k = z; k > 0; k -= lb(k)) {
                s += sum[i][j][k];
            }
        }
    }
    return s;
}

void update(int x, int y, int z) {
    //add one
    //++x, ++y, ++z;
    for (int i = x; i < MAX; i += lb(i)) {
        for (int j = y; j < MAX; j += lb(j)) {
            for (int k = z; k < MAX; k += lb(k)) {
                sum[i][j][k] ^= 1;
            }
        }
    }
}

int main() {
    int n, m, op;
    while (~scanf(" %d %d", &n, &m)) {
        memset(sum, 0, sizeof(sum));
        int x1, y1, z1, x2, y2, z2;
        while (m--) {
            scanf(" %d %d %d %d", &op, &x1, &y1, &z1);
            if (op == 0) {
                printf("%d\n", query(x1, y1, z1) & 1);
            } else {
                scanf(" %d %d %d", &x2, &y2, &z2);
                update(x1, y1, z1); //add
                update(x2 + 1, y1, z1); //sub
                update(x1, y2 + 1, z1); //sub
                update(x1, y1, z2 + 1); //sub
                update(x2 + 1, y2 + 1, z1); //add
                update(x2 + 1, y1, z2 + 1); //add
                update(x1, y2 + 1, z2 + 1); //add
                update(x2 + 1, y2 + 1, z2 + 1); //sub
            }
        }
    }
    return 0;
}