题意
二维书架上原本有1本书
T组样例,四种操作,search,add,del,move
书架上的书小于del的值时,取完该格子的书即可
查询操作用二维的容斥推
由于格子取值可以等于0,故下标都+1
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 1e3 + 5;
const int MOD = 1e9 + 9;
const double pi = 3.1415926;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define F(i, l, r) for(int i = l;i <= (r);++i)
#define RF(i, l, r) for(int i = l;i >= (r);--i)
int sum[N << 1][N << 1], p[N][N];
int lowbit(int x) {return x & (-x);}
void change(int x, int y, int num)
{
for(int i = x;i <= N;i += lowbit(i))
for(int j = y;j <= N;j += lowbit(j))
sum[i][j] += num;
}
int query(int x, int y)
{
int ret = 0;
for(int i = x;i;i -= lowbit(i))
for(int j = y;j;j -= lowbit(j))
ret += sum[i][j];
return ret;
}
int main()
{
//freopen("in.txt", "r", stdin);
int T;cin >> T;
for(int cas = 1;cas <= T;++cas)
{
cout << "Case " << cas << ":" << endl;
for(int i = 1;i < N;++i)
for(int j = 1;j < N;++j)
{sum[i][j] = lowbit(i) * lowbit(j);p[i][j] = 1;}
int Q;cin >> Q;
while(Q--)
{
string s;int a, b, c, d, e;
cin >> s;
if(s == "A")
{
cin >> a >> b >> c;
a++;b++;p[a][b] += c;
change(a, b, c);
}
else if(s == "S")
{
cin >> a >> b >> c >> d;
a++;b++;c++;d++;
if(a > c) swap(a, c);
if(b > d) swap(b, d);
cout << query(c, d) - query(a - 1, d) - query(c, b - 1) + query(a - 1, b - 1) << endl;
}
else if(s == "D")
{
cin >> a >> b >> c;
a++;b++;c = min(c, p[a][b]);p[a][b] -= c;
change(a, b, -c);
}
else
{
cin >> a >> b >> c >> d >> e;
a++;b++;c++;d++;
e = min(e, p[a][b]);p[a][b] -= e;p[c][d] += e;
change(a, b, -e);change(c, d, e);
}
}
}
return 0;
}