AcWing 阿巴阿巴(没买). 离散化
原题链接
简单
作者:
未名湖畔的梦
,
2021-04-16 16:39:16
,
所有人可见
,
阅读 203
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5+5;
typedef pair<int,int>PII;
int n, m, a[N], s[N];
vector<int>alls;
vector<PII>add, query;
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(){
scanf("%d%d",&n,&m);
for(int i = 0; i < n; i++){
int x, c;
scanf("%d%d",&x,&c);
add.push_back({x,c});
alls.push_back(x);
}
for(int i = 0; i < m; i++){
int l ,r ;
scanf("%d%d",&l,&r);
query.push_back({l,r});
alls.push_back(l);
alls.push_back(r);
}
sort(alls.begin(), alls.end());
alls.erase(unique(alls.begin(), alls.end()), alls.end());
for(auto i : add){
int x = find(i.first);
a[x] += i.second;
}
for(int i = 1; i <= alls.size(); i++) s[i] = s[i-1] + a[i];
for(auto i : query){
int l = find(i.first), r = find(i.second);
cout<<s[r] - s[l-1]<<endl;
}
return 0;
}