# 离散化算法总结

1. 原数组中可能有重复元素，需要去重。

2. 如何算出离散化后的值（离散化后保序，使用二分）。

https://www.acwing.com/problem/content/804/

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

typedef pair<int, int> PII;

const int N = 300010;

int n, m;
int a[N], s[N];

vector<int> alls;

int find(int x)
{
int l = 0, r = alls.size() - 1;
while (l < r)
{
int mid = l + r >> 1;
if (alls[mid] >= x) r = mid;
else l = mid + 1;
}
return r + 1;
}

int main()
{
cin >> n >> m;
for (int i = 0; i < n; i++)
{
int x, c;
cin >> x >> c;

alls.push_back(x);
}

for (int i = 0; i < m; i++)
{
int l, r;
cin >> l >> r;
query.push_back({l, r});

alls.push_back(l);
alls.push_back(r);
}

// 去重
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());

// 处理插入
{
int x = find(item.first);
a[x] += item.second;
}

// 预处理前缀和
for (int i = 1; i <= alls.size(); i++) s[i] = s[i - 1] + a[i];

// 处理询问
for (auto item : query)
{
int l = find(item.first), r = find(item.second);
cout << s[r] - s[l - 1] << endl;
}

return 0;
}

vector<int>::iterator unique(vector<int>& a)
{
int j = 0;
for (int i = 0; i < a.size(); i++)
if (!i || a[i] != a[i - 1])
a[j++] = a[i];
// a[0] ~ a[j - 1] 所有a中不重复的数

return a.begin() + j;
}

posted @ 2023-12-02 00:09  ykycode  阅读(38)  评论(0编辑  收藏  举报