/*
覺得真是…經歷了高考啥都忘了(其實還是當初學得不踏實
現在一點一點再重新來補吧(過了一年再來說這話的我
*/

參考資料及模板

poj題目小集

看上去是個二維的題，事實上因為讀入資料是按序排列的，所以可以直接轉化成一維來做，就是個裸的單點修改區間查詢的樹狀陣列了。
也給了我們啟示，有時對讀入資料排序處理一下，就能收到很好的效果。
還有要注意的就是，樹狀陣列維護的區間下標要從1開始，這道題WA了幾次才注意到。

Code:

#include <cstdio>
#include <cstring>
#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;
}

兩道用 樹狀陣列 來解決 逆序對 的問題。
以前一直都只知道用歸併來寫的我實在是…個人覺得寫起來比歸併好寫多了。
第一題就是裸的。
第二題還是先通過排序處理一下，做起來就方便不少。

poj 2299 Code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#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 <cstdio>
#include <cstring>
#include <algorithm>
#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;
}

要求一個序列中單調增或者單調減的三元組的個數總和。
寫起來和逆序對差不多，從左起插一遍，右起插一遍，得到每個數左邊比它大的、比它小的；右邊比它大的、比它小的數的個數。其實就是列舉中間那個數，最後乘一乘就是答案。
這道題一個注意點就是題目中的 distinct integers，一開始沒注意到，寫得又稍微複雜了些，記錄了每個數迄今出現的次數 cnt，然後再搞一搞啥的。其實也沒啥…。

Code:

#include <cstdio>
#include <cstring>
#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;
}

裸的二維樹狀陣列
一開始還是用while在那裡寫（太重的Pascal痕跡= =
後來看了模板改成優雅的for了…

Code:

#include <cstdio>
#include <cstring>
#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;
}