离散化
模板
vector<int> alls; // 存储所有待离散化的值
sort(alls.begin(), alls.end()); // 将所有值排序
alls.erase(unique(alls.begin(), alls.end()), alls.end()); // 去掉重复元素
// 二分求出x对应的离散化的值
int find(int x) // 找到第一个大于等于x的位置
{
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; // 映射到1, 2, ...n
}
模板题——区间和
代码
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 300010; //n + 2m的范围是三十万
int n, m, a[N], s[N];
vector<int> alls;
vector<PII> add, query;
int find(int x)
{
int l = 1, r = alls.size();
while(l < r)
{
int mid = (l + r) / 2;
if(alls[mid] >= x) r = mid;
else l = mid + 1;
}
return r + 1;
}
int main()
{
scanf("%d%d", &n, &m);
while(n--)
{
int x, c;
scanf("%d%d", &x, &c);
alls.push_back(x);
add.push_back({x, c});
}
while(m--)
{
int l, r;
scanf("%d%d", &l, &r);
alls.push_back(l);
alls.push_back(r);
query.push_back({l, r});
}
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());
for(auto item : add)
{
int x = find(item.first);
a[x] += item.second;
}
for(int i = 1; i <= alls.size(); i++)
s[i] = s[i-1] + a[i];
for(auto item : query)
{
int l = find(item.first), r = find(item.second);
printf("%d\n", s[r] - s[l-1]);
}
return 0;
}