「题解」「数据结构」树状数组3

接着上文说

---

离散化

for (int i = 1; i <= n; i ++) {    
	cin >> a[i].val;    
	a[i].id = i;
}
sort(a + 1, a + n + 1);           //定义结构体时按val从小到大重载
for (int i = 1; i <= n; i ++)    
	b[a[i].id] = i;                    //将a[i]数组映射成更小的值,b[i]就是a[i]对应的rank(顺序)值

for (int i = 1; i <= n; i ++) {	
	scanf("%d",&num[i]);	
	lsh[i] = num[i];	//复制一份原数组
}
sort(lsh + 1, lsh + n + 1); //排序，unique虽有排序功能，但交叉数据排序不支持，所以先排序防止交叉数据
//cnt就是排序去重之后的长度
cnt = unique(lsh + 1, lsh + n + 1) - lsh - 1; //unique返回去重之后最后一位后一位地址 - 数组首地址 - 1
for (int i = 1; i <= n; i++)	
      num[i] = lower_bound(lsh + 1, lsh + cnt + 1, num[i]) - lsh;

代码+解释

#include 
#include 
using namespace std;

const int M = 1e3 + 5;
int a[M], dis[M], bit[M];
int n;

void Read() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
	}
}

void Discretetion() {
	for (int i = 1; i <= n; i ++) {
		dis[i] = a[i];
	}
	sort(dis + 1, dis + 1 + n);
	int len = unique(dis + 1, dis + 1 + n) - dis - 1;
	for (int i = 1; i <= n; i ++) {
		a[i] = lower_bound(dis + 1, dis + 1 + n, a[i]) - dis;
	}
}

int lowbit(int x) { return x & (- x); }

void update(int k) {
	for (int i = k; i <= n; i += lowbit(i)) {
		bit[i] ++;				//从之前的修改值，变为增加一次
	}
}

int Sum(int k) {
	int tot = 0;
	for (int i = k; i > 0; i -= lowbit(i)) {
		tot += bit[i];
	}
	return tot;
}

void work() {
	int ans = 0;
	for (int i = 1; i <= n; i ++) {
		update(a[i]);				//因为逆序对需要保证下标按顺序，所以必须放一个，求一次
		ans += i - Sum(a[i]);				//累加
	}
	printf("%d", ans);
}

int main() {
	Read();				//输入
	Discretetion();				//离散化
	work();				//输出
	return 0;
}