POJ 1195 Mobile phones(裸的二维树状数组)
http://poj.org/problem?id=1195
题意:给出一个矩阵,给某个格子加/减一个数,就某个子矩阵的和,1024*1024的范围,二维的树状数组
子矩阵(x1,y1,x2,y2)(左下角的点和右上角的点)的和ans=sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1)
代码:
#include<iostream> #include<cstdio> #include<cstring> #include<string> #define nMAX 1050 using namespace std; int c[nMAX][nMAX],n; int lowbit(int x) { return x&(-x); } void add(int i,int j,int w) { int temp; while(i<=n) { temp=j; while(temp<=n) { c[i][temp]+=w; temp+=lowbit(temp); } i+=lowbit(i); } } int sum(int i,int j)//与一维的稍有不同 { int ans=0,temp; while(i>=1) { temp=j; while(temp>=1) { ans+=c[i][temp]; temp-=lowbit(temp); } i-=lowbit(i); } return ans; } int main() { int i,j,k,f,commend; while(~scanf("%d",&commend)) { if(commend==3)break; if(commend==0) { scanf("%d",&n); memset(c,0,sizeof(c)); } if(commend==1) { scanf("%d%d%d",&i,&j,&k);//下表必须从1,1开始,因为lowbit(0)=0死循环 add(i+1,j+1,k); } if(commend==2) { scanf("%d%d%d%d",&i,&j,&k,&f); i++,j++,k++,f++; int ans=sum(k,f)-sum(i-1,f)-sum(k,j-1)+sum(i-1,j-1); printf("%d\n",ans); } } return 0; }