二维树状数组入门题 poj2642Stars
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题目连接:Stars
题解:把一维的的树状数组扩展到二维就行,复杂度为o(mlog^2n)
#include<bits/stdc++.h> #include<set> #include<cstdio> #include<iomanip> #include<iostream> #include<string> #include<cstring> #include<algorithm> #define pb push_back #define ll long long #define fi first #define se second #define PI 3.14159265 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define eps 1e-7 #define pii pair<int,int> typedef unsigned long long ull; const int mod=1e3+5; const ll inf=0x3f3f3f3f3f3f3f; const int maxn=1e3+5; using namespace std; int n,m,a[maxn][maxn]; bool vis[maxn][maxn]; int lower_bit(int x){return x&(-x);} void add(int x,int y,int val) { for(int i=x;i<maxn;i+=lower_bit(i)) { for(int j=y;j<maxn;j+=lower_bit(j)) { a[i][j]+=val; } } } int get_sum(int x,int y) { int ans=0; while(x>0) { for(int j=y;j>0;j-=lower_bit(j)) { ans+=a[x][j]; } x-=lower_bit(x); } return ans; } int main() { // ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); // cin>>n; scanf("%d",&n); while(n--) { int x,y,x2,y2;char op[10]; // cin>>op; scanf("%s",&op); if(op[0]==‘B‘) { //cin>>x>>y; scanf("%d %d",&x,&y);x++;y++; if(!vis[x][y])add(x,y,1),vis[x][y]=1; } else if(op[0]==‘D‘) { //cin>>x>>y; scanf("%d %d",&x,&y);x++,y++; if(vis[x][y])add(x,y,-1),vis[x][y]=false; } else { //cin>>x>>x2>>y>>y2; scanf("%d %d %d %d",&x,&x2,&y,&y2);x++;x2++,y2++;y++; if(x>x2)swap(x,x2);if(y>y2)swap(y,y2); printf("%d ",get_sum(x2,y2)-get_sum(x-1,y2)-get_sum(x2,y-1)+get_sum(x-1,y-1)); // cout<<get_sum(x2,y2)-get_sum(x-1,y2)+get_sum(x2,y-1)+get_sum(x-1,y-1)<<endl; } } return 0; }
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