树状数组模板及poj几道简单题

/*
觉得真是…经历了高考啥都忘了(其实还是当初学得不踏实
现在一点一点再重新来补吧(过了一年再来说这话的我
*/

参考资料及模板

http://blog.csdn.net/lawrence_jang/article/details/8054173 ——Lawrence_Jang
http://blog.csdn.net/qq_21841245/article/details/43956633 ——MoeO3

poj题目小集

poj 2352

看上去是个二维的题，事实上因为读入数据是按序排列的，所以可以直接转化成一维来做，就是个裸的单点修改区间查询的树状数组了。
也给了我们启示，有时对读入数据排序处理一下，就能收到很好的效果。
还有要注意的就是，树状数组维护的区间下标要从1开始，这道题WA了几次才注意到。

Code:

#include 
#include 
#define maxn 32001
#define maxm 15010
inline int lowbit(int x) { return x & (-x); }
int n, level[maxm], c[maxn + 10];
void update(int x) {
    while (x <= maxn) {
        ++c[x];
        x += lowbit(x);
    }
}
int query(int x) {
    int ret = 0;
    while (x) {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}
void work() {
    memset(level, 0, sizeof(level));
    memset(c, 0, sizeof(c));
    for (int i = 0; i < n; ++i) {
        int x, y;
        scanf("%d%d", &x, &y);
        update(++x);
        ++level[query(x)];
    }
    for (int i = 1; i <= n; ++i) printf("%d\n", level[i]);
}
int main() {
    while (scanf("%d", &n) != EOF) work();
    return 0;
}

poj 2299 & poj 3067

两道用 树状数组 来解决 逆序对 的问题。
以前一直都只知道用归并来写的我实在是…个人觉得写起来比归并好写多了。
第一题就是裸的。
第二题还是先通过排序处理一下，做起来就方便不少。

poj 2299 Code:

#include 
#include 
#include 
#define maxn 500010
using namespace std;
typedef long long LL;
struct node {
    int val, p;
    bool operator < (const node& nd) const { return val < nd.val; }
}a[maxn];
int b[maxn], c[maxn], n;
inline int lowbit(int x) { return x & (-x); }
int query(int x) {
    int ret = 0;
    while (x) {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}
void add(int x) {
    while (x <= n) {
//        printf("%d\n", x);
        ++c[x];
        x += lowbit(x);
    }
}
void work() {
    memset(c, 0, sizeof(c));
    for (int i = 1; i <= n; ++i) { scanf("%d", &a[i].val); a[i].p = i; }
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= n; ++i) b[a[i].p] = i;
    LL sum = 0;
    for (int i = 1; i <= n; ++i) {
        sum += i - 1 - query(b[i]);
        add(b[i]);
    }
    printf("%lld\n", sum);
}
int main() {
    freopen("in.txt", "r", stdin);
    while (scanf("%d", &n) != EOF && n) work();
    return 0;
}

poj 3067 Code:

#include 
#include 
#include 
#define maxn 1010
#define maxk 1000010
int c[maxn], kas, n, m, k;
using namespace std;
typedef long long LL;
struct Edge {
    int x, y;
}e[maxk];
bool cmp(Edge a, Edge b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
inline int lowbit(int x) { return x & (-x); }
int query(int x) {
    int ret = 0;
    while (x) {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}
void add(int x) {
    while (x <= m) {
        ++c[x];
        x += lowbit(x);
    }
}
void work() {
    memset(c, 0, sizeof(c));
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < k; ++i) scanf("%d%d", &e[i].x, &e[i].y);
    sort(e, e + k, cmp);
    LL sum = 0;
    for (int i = 0; i < k; ++i) {
        sum += i - query(e[i].y);
        add(e[i].y);
    }
    printf("Test case %d: %lld\n", ++kas, sum);
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}

poj 3928

要求一个序列中单调增或者单调减的三元组的个数总和。
写起来和逆序对差不多，从左起插一遍，右起插一遍，得到每个数左边比它大的、比它小的；右边比它大的、比它小的数的个数。其实就是枚举中间那个数，最后乘一乘就是答案。
这道题一个注意点就是题目中的 distinct integers，一开始没注意到，写得又稍微复杂了些，记录了每个数迄今出现的次数 cnt，然后再搞一搞啥的。其实也没啥…。

Code:

#include 
#include 
#define maxn 100000
#define maxm 20010
int a[maxm], c[maxn + 10], cnt[maxn + 10], le1[maxm], le2[maxm], gr1[maxm], gr2[maxm];
typedef long long LL;
inline int lowbit(int x) { return x & (-x); }
void add(int x) {
    while (x <= maxn) {
        ++c[x];
        x += lowbit(x);
    }
}
int query(int x) {
    int ret = 0;
    while (x) {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}
void work() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    memset(c, 0, sizeof(c));
    for (int i = 0; i < n; ++i) {
        int x = a[i];
        le1[i] = query(x);
        gr1[i] = i - le1[i];
        add(x);
    }
    memset(c, 0, sizeof(c));
    for (int i = n - 1; i >= 0; --i) {
        int x = a[i];
        le2[i] = query(x);
        gr2[i] = (n - 1 - i) - le2[i];
        add(x);
    }
//    for (int i = 1; i < n - 1; ++i) {
//        printf("%d %d %d %d\n", le1[i], le2[i], gr1[i], gr2[i]);
//    }
    LL ans = 0;
    for (int i = 1; i < n - 1; ++i) {
        ans += le1[i] * gr2[i] + le2[i] * gr1[i];
    }
    printf("%lld\n", ans);
}
int main() {
    freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) work();
    return 0;
}

poj1195

裸的二维树状数组
一开始还是用while在那里写（太重的Pascal痕迹= =
后来看了模板改成优雅的for了…

Code:

#include 
#include 
#define maxn 1030
int c[maxn][maxn], s, n;
inline int lowbit(int x) { return x & (-x); }
void update(int x, int y, int a) {
    for (int i = x; i <= n; i += lowbit(i)) {
        for (int j = y; j <= n; j += lowbit(j)) {
            c[i][j] += a;
        }
    }
}
int query(int x, int y) {
    if (x == 0 || y == 0) return 0;
    int ret = 0;
    for (int i = x; i; i -= lowbit(i)) {
        for (int j = y; j; j -= lowbit(j)) {
            ret += c[i][j];
        }
    }
    return ret;
}
void work() {
    memset(c, 0, sizeof(c));
    while (scanf("%d", &s)) {
        if (s == 3) return;
        if (s == 1) {
            int x, y, a;
            scanf("%d%d%d", &x, &y, &a);
            ++x; ++y;
            update(x, y, a);
        }
        else {
            int l, b, r, t;
            scanf("%d%d%d%d", &l, &b, &r, &t);
            ++r; ++t;
            printf("%d\n", query(r, t) - query(l, t) - query(r, b) + query(l, b));
        }
    }
}
int main() {
    freopen("in.txt", "r", stdin);
    while (scanf("%d%d", &s, &n) != EOF) work();
    return 0;
}