【BZOJ-1452】Count 树状数组 套 树状数组

1452: [JSOI2009]Count

Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 1769 Solved: 1059
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Description

Input

Output

Sample Input

Sample Output

1

2

HINT

Source

Solution

忽略标题的说法...撞壁用的....

简单题，裸树状数组套树状数组

颜色数目$c<=100$很小，考虑对于每种颜色单独进行处理

那么对于每个颜色单独开一个二维树状数组维护

另开一个二维数组记录下当前各个位置的颜色

1操作，将原有颜色树状数组中的$（x,y）$点置成0，在新颜色树状数组中置成1，并更新颜色的状态

2操作，区间查询，应用前缀和，进行一下加减即可得到当前子矩阵和

Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int read()
{
int x=0,f=1; char ch=getchar();
while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n,m,q;int zt[310][310];
struct Treenode
{
int tree[310][310];
Treenode(){memset(tree,0,sizeof(tree));}
int lowbit(int x){return x&-x;}
void change(int x,int y,int del)
{
for (int i=x; i<=n; i+=lowbit(i))
for (int j=y; j<=m; j+=lowbit(j))
tree[i][j]+=del;
}
int query(int x,int y)
{
int ans=0;
for (int i=x; i; i-=lowbit(i))
for (int j=y; j; j-=lowbit(j))
ans+=tree[i][j];
return ans;
}
int ask(int x1,int y1,int x2,int y2)
{
return query(x2,y2)-query(x2,y1-1)-query(x1-1,y2)+query(x1-1,y1-1);
}
}color[110];

int main()
{
n=read(); m=read();
for (int i=1; i<=n; i++)
for (int col,j=1; j<=m; j++)
col=read(),color[col].change(i,j,1),zt[i][j]=col;
q=read();
for (int i=1; i<=q; i++)
{
int opt=read(),x1,y1,x2,y2,col;
if (opt==1)
{
x1=read(),y1=read(),col=read();
color[zt[x1][y1]].change(x1,y1,-1);
zt[x1][y1]=col;
color[col].change(x1,y1,1);
continue;
}
x1=read(),x2=read(),y1=read(),y2=read(),col=read();
printf("%d\n",color[col].ask(x1,y1,x2,y2));
}
return 0;
}

5分钟不到，看、想、写、1A系列....