【bzoj 1452】[JSOI2009]Count(二维树状数组)
1452: [JSOI2009]Count
Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 2094 Solved: 1232
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Description
Input
Output
Sample Input
Sample Output
1
2
2
HINT
Source
【题解】【树状数组】
【二维树状数组,不要想太多,只要把一维改成二维,给每种权值建一棵树状数组(权值数不超过100种),然后正常做就行】
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int tree[310][310][110];
int a[310][310],n,m,q;
inline void add(int x,int y,int num,int val)
{
for (int i=x;i<=n;i+=i&(-i))
for (int j=y;j<=m;j+=j&(-j))
tree[i][j][num]+=val;
}
inline int ask(int x,int y,int num)
{
int sum=0;
for (int i=x;i>=1;i-=i&(-i))
for (int j=y;j>=1;j-=j&(-j))
sum+=tree[i][j][num];
return sum;
}
int main()
{
int i,j;
scanf("%d%d",&n,&m);
for (i=1;i<=n;++i)
for (j=1;j<=m;++j)
{
scanf("%d",&a[i][j]);
add(i,j,a[i][j],1);
}
scanf("%d",&q);
for (i=1;i<=q;++i)
{
int num;
scanf("%d",&num);
if (num==1)
{
int x,y,val;
scanf("%d%d%d",&x,&y,&val);
add(x,y,a[x][y],-1);
add(x,y,val,1);
a[x][y]=val;
}
else
{
int x1,x2,y1,y2,val,ans;
scanf("%d%d%d%d%d",&x1,&x2,&y1,&y2,&val);
int ans1=ask(x2,y2,val);
int ans2=ask(x1-1,y2,val);
int ans3=ask(x2,y1-1,val);
int ans4=ask(x1-1,y1-1,val);
ans=ans1-ans2-ans3+ans4;
printf("%d\n",ans);
}
}
}
既然无能更改,又何必枉自寻烦忧