【离散化】AcWing802. 区间和

AcWing802.区间和

题解

本题要是一个很长的数轴，已经超过了数组的长度1e6，故我们需要使用离散化压缩空间

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

using namespace std;

typedef pair<int,int> PII;

const int N = 3e5+10; //n输入的点 与m个查询区间的l,r

vector<int> alls;
vector<PII> add, query;
int a[N], s[N];

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

int main()
{
    int n, m, x, y;
    scanf("%d%d",&n,&m);
    for(int i = 0; i < n; ++i)
    {
        scanf("%d%d",&x,&y);
        alls.push_back(x);
        add.push_back({x,y});
    }
    
    for(int i = 0; i < m; ++i)
    {
        scanf("%d%d",&x,&y);
        alls.push_back(x),alls.push_back(y);
        query.push_back({x,y}); //把查询也加入进去才可找到的查询l，r的离散化映射
    }
    
    sort(alls.begin(), alls.end());
    alls.erase(unique(alls.begin(),alls.end()), alls.end());
    
    for(auto item : add)
    {
        int i = find(item.first);
        a[i] += 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：AcWing802,int,back,mid,离散,push,alls,区间,include
来源： https://www.cnblogs.com/czy-algorithm/p/16252277.html