离散化
题目链接：https://www.acwing.com/problem/content/description/804/
思路：
离散化实质是一种映射
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int N=3e5+10; 4 typedef pair<int,int> PII; 5 vector<int> idx; 6 vector<PII> add,query; 7 int a[N],d[N]; 8 int n,m; 9 int find(int x) 10 { 11 int l=0,r=idx.size()-1; 12 while(l<r){ 13 int mid=l+r>>1; 14 if(idx[mid]>=x) r=mid; 15 else l=mid+1; 16 } 17 return r+1; 18 } 19 int main() 20 { 21 cin>>n>>m; 22 while(n--){ 23 int x,c; 24 cin>>x>>c; 25 idx.push_back(x); 26 add.push_back({x,c}); 27 } 28 while(m--){ 29 int l,r; 30 cin>>l>>r; 31 query.push_back({l,r}); 32 idx.push_back(l); 33 idx.push_back(r); 34 } 35 sort(idx.begin(),idx.end()); 36 idx.erase(unique(idx.begin(),idx.end()),idx.end()); 37 for(int i=0;i<add.size();i++){ 38 int x=find(add[i].first); 39 a[x]+=add[i].second; 40 } 41 for(int i=1;i<=idx.size();i++){ 42 d[i]=d[i-1]+a[i]; 43 } 44 for(int i=0;i<query.size();i++){ 45 int l=find(query[i].first),r=find(query[i].second); 46 cout<<d[r]-d[l-1]<<endl; 47 } 48 return 0; 49 }
