BZOJ1452Count

1452: [JSOI2009]Count

Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 1687 Solved: 1000
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Description

Input

Output

Sample Input
3 3
1 2 3
3 2 1
2 1 3
3
2 1 2 1 2 1
1 2 3 2
2 2 3 2 3 2
Sample Output
1
2
HINT

Source
JSOI2009Day1
建c个二维树状数组，读入的矩阵当做c，先加1，单点更新的时候先把加的1减去，用更新的c更新矩阵，查询个数的时候直接二维树状数组求和即可。。
只不过多一个for循环。。变量名重复错误弄了我好久QAQ
感谢hzwer学长。。

#include<cstdio>using namespace std;int n,m,k,a[301][301],f[101][301][301];int read(){    int x=0,w=1;    char ch=getchar();    while (ch<'0' || ch>'9')      {        if (ch=='-')          w=-1;        ch=getchar();      }    while (ch>='0' && ch<='9')      {        x=x*10+ch-'0';        ch=getchar();      }    return x*w;}int lowbit(int x){    return x&(-x);}void add(int i,int j,int c,int value){    int x,y;    for (x=i;x<=n;x+=lowbit(x))      for (y=j;y<=m;y+=lowbit(y))        f[c][x][y]+=value;}int query(int i,int j,int c){    int ans=0,x,y;    for (x=i;x>=1;x-=lowbit(x))      for (y=j;y>=1;y-=lowbit(y))        ans+=f[c][x][y];    return ans;}int main(){    int i,j,p,x1,y1,x2,y2,c;    n=read();    m=read();    for (i=1;i<=n;i++)      for (j=1;j<=m;j++)        {            a[i][j]=read();            add(i,j,a[i][j],1);        }    k=read();    for (i=1;i<=k;i++)      {        p=read();        if (p==1)          {            x1=read();            y1=read();            c=read();            add(x1,y1,a[x1][y1],-1);            a[x1][y1]=c;            add(x1,y1,a[x1][y1],1);          }        if (p==2)          {            x1=read();            x2=read();            y1=read();            y2=read();            c=read();            printf("%d\n",query(x2,y2,c)+query(x1-1,y1-1,c)-query(x1-1,y2,c)-query(x2,y1-1,c));          }      }    return 0;}