离散化

• 模板
• 模板题——区间和
• • 代码

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;
}

标签：end,int,基础,mid,back,离散,算法,alls,find
来源： https://blog.csdn.net/cloudy_sqyin/article/details/119464921