【树状数组（二维）】poj 1195 Mobile phones（外：hdu 2642 Stars）

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二维插点，求区间：

4（sum(x2,y2) + 1（sum(x1-1,y1-1)）- 2（sum(x1-1,y2)）-3（sum(x2,y1-1)））

http://poj.org/problem?id=1195

分析:裸二维树状数组

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int NM=1030;__int64 a[NM][NM];int lowbit(int x){return x&(-x);}void add(int x,int y,int t){for(int i=x;i<NM;i+=lowbit(i))for(int j=y;j<NM;j+=lowbit(j)){if(a[i][j]+t<0) a[i][j]=0;  //else a[i][j]+=t;}}__int64 sum(int x,int y){__int64 ans=0;for(int i=x;i>0;i-=lowbit(i))for(int j=y;j>0;j-=lowbit(j))ans+=a[i][j];return ans;}int main(){int x1,y1,x2,y2,t,n,temp;scanf("%d%d",&t,&n);while(scanf("%d",&t)){if(t==3) break;else if(t==1){scanf("%d%d%d",&x1,&y1,&temp);x1++,y1++;add(x1,y1,temp);}else if(t==2){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++,y1++,x2++,y2++;printf("%I64d\n",sum(x2,y2)+sum(x1-1,y1-1)-sum(x1-1,y2)-sum(x2,y1-1));}}return 0;}

http://acm.hdu.edu.cn/showproblem.php?pid=2642

分析:裸二维树状数组

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int NM=1005;int a[NM][NM];bool f[NM][NM];int lowbit(int x){return x&(-x);}void add(int x,int y,int t){for(int i=x;i<NM;i+=lowbit(i))for(int j=y;j<NM;j+=lowbit(j))a[i][j]+=t;}int sum(int x,int y)    //从1开始算{int ans=0;for(int i=x;i>0;i-=lowbit(i))for(int j=y;j>0;j-=lowbit(j))ans+=a[i][j];return ans;}int main(){int x1,y1,x2,y2,x,y,T;char cc[5];while(scanf("%d",&T)!=EOF){memset(f,0,sizeof(f));  //开始时灯都是灭的memset(a,0,sizeof(a));while(T--){scanf("%s",cc);if(cc[0]=='D'){scanf("%d%d",&x,&y);x++,y++;if(!f[x][y]) continue;f[x][y]=false;add(x,y,-1);}else if(cc[0]=='B')  //bright{scanf("%d%d",&x,&y);x++,y++;if(f[x][y]) continue;f[x][y]=true;add(x,y,1);}else{scanf("%d%d%d%d",&x1,&x2,&y1,&y2);x1++,y1++,x2++,y2++;if(x1>x2) swap(x1,x2);if(y1>y2) swap(y1,y2);//注意边界printf("%d\n",sum(x2,y2)+sum(x1-1,y1-1)-sum(x2,y1-1)-sum(x1-1,y2)); }}}return 0;}

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