BZOJ 1452: [JSOI2009]Count(二维BIT)

为每一个权值开一个二维树状数组.

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

 

#define rep(i, n) for(int i = 0; i < n; ++i)

#define Rep(i ,n) for(int i = 1; i <= n; ++i)

 

using namespace std;

 

const int maxn = 300 + 5;

 

#define lowbit(x) ((x) & -(x))

 

int n, m;

 

struct BIT {

int b[maxn][maxn];

BIT() { memset(b, 0, sizeof b); }

void add(int x, int y, int v) {

for(int i = x; i <= n; i += lowbit(i))

   for(int j = y; j <= m; j += lowbit(j))

       b[i][j] += v;

}

int sum(int x, int y) {

int ans = 0;

for(int i = x; i > 0; i -= lowbit(i))

   for(int j = y; j > 0; j -= lowbit(j))

       ans += b[i][j];

return ans;

}

};

 

const int maxc = 100 + 5;

 

BIT g[maxc];

int M[maxn][maxn];

 

int main() {

freopen("test.in", "r", stdin);

cin >> n >> m;

Rep(i, n)

   Rep(j, m) {

    scanf("%d", &M[i][j]);

    g[M[i][j]].add(i, j, 1);

   }

int q;

cin >> q;

while(q--) {

int op;

scanf("%d", &op);

if(op == 1) {

int x, y, v;

scanf("%d%d%d", &x, &y, &v);

g[M[x][y]].add(x, y, -1);

g[M[x][y] = v].add(x, y, 1);

} else {

int x1, x2, y1, y2, v;

scanf("%d%d%d%d%d", &x1, &x2, &y1, &y2, &v);

x1--; y1--;

printf("%d\n", g[v].sum(x2, y2) + g[v].sum(x1, y1) - g[v].sum(x1, y2) - g[v].sum(x2, y1));

}

}

return 0;

}

  

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1452: [JSOI2009]Count

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 1384  Solved: 814
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Description

Input

Output

Sample Input

Sample Output

1
2

HINT

Source

JSOI2009Day1