【传送门：BZOJ1452】
简要题意：
给出一个n*m的矩阵，共有两种操作：
1 x y c将第x行第y列的数改为c
2 x1 x2 y1 y2 c求出第x1行第y1列到第x2行第y2列值为c的格子数
题解：
第一次写二维树状数组，和一维差不多
a[x][y][c]表示第1行第1列到第x行第y列值为c的格子数
然后维护树状数组，求值时，要用矩阵的思想来求
参考代码：
#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;int a[310][310][110];int n,m;int lowbit(int x){return x&-x;}void change(int x,int y,int c,int d){ ???int t=y; ???while(x<=n) ???{ ???????y=t; ???????while(y<=m) ???????{ ???????????a[x][y][c]+=d; ???????????y+=lowbit(y); ???????} ???????x+=lowbit(x); ???}}int getsum(int x,int y,int c){ ???int ans=0,t=y; ???while(x!=0) ???{ ???????y=t; ???????while(y!=0) ???????{ ???????????ans+=a[x][y][c]; ???????????y-=lowbit(y); ???????} ???????x-=lowbit(x); ???} ???return ans;}int map[310][310];int main(){ ???scanf("%d%d",&n,&m); ???memset(a,0,sizeof(a)); ???for(int i=1;i<=n;i++) ???{ ???????for(int j=1;j<=m;j++) ???????{ ???????????scanf("%d",&map[i][j]); ???????????change(i,j,map[i][j],1); ???????} ???} ???int q; ???scanf("%d",&q); ???while(q--) ???{ ???????int t; ???????scanf("%d",&t); ???????if(t==1) ???????{ ???????????int x,y,c; ???????????scanf("%d%d%d",&x,&y,&c); ???????????change(x,y,map[x][y],-1); ???????????map[x][y]=c; ???????????change(x,y,c,1); ???????} ???????else ???????{ ???????????int x1,x2,y1,y2,c; ???????????scanf("%d%d%d%d%d",&x1,&x2,&y1,&y2,&c); ???????????printf("%d\n",getsum(x2,y2,c)-getsum(x1-1,y2,c)-getsum(x2,y1-1,c)+getsum(x1-1,y1-1,c)); ???????} ???} ???return 0;}BZOJ1452: [JSOI2009]Count
原文地址：https://www.cnblogs.com/Never-mind/p/8621365.html