二维树状数组模板(区间修改+区间查询)

二维树状数组模板(区间修改+区间查询)

例题:JOIOI上帝造题的七分钟
一共两种操作:

\(L\ x_1\ y_1\ x_2\ y_2\ d\):把\((x_1,y_1)\)\((x_2,y_2)\)这个矩形内所有元素加\(d\)
\(k\ x_1\ y_1\ x_2\ y_2\):查询\((x_1,y_1)\)\((x_2,y_2)\)这个矩形内所有元素的和。

代码如下:

#include
#define RG register
#define IL inline
#define _ 2050
#define ll long long
//#define ll int
using namespace std;

ll n , m , c1[_][_],c2[_][_],c3[_][_],c4[_][_] ;

IL void upt(ll x , ll y , ll dt){
    for(RG int i = x; i <= n; i += (i & -i) )
        for(RG ll j = y; j <= m; j += (j & -j)){
            c1[ i ][ j ] += dt ;
            c2[ i ][ j ] += dt * y;
            c3[ i ][ j ] += dt * x;
            c4[ i ][ j ] += dt * x * y;
        }
}
IL ll calc(ll x,ll y){
    RG ll res = 0;
    for(RG ll i = x; i > 0; i -= (i & -i))
        for(RG ll j = y; j > 0; j -= (j & -j)){
            res = res
                + (x + 1) * (y + 1) * c1[ i ][ j ]
                - (x + 1) * c2[ i ][ j ]
                - (y + 1) * c3[ i ][ j ]
                + c4[ i ][ j ] ; 
        }
    return res;
}

IL void add(ll X1,ll Y1,ll X2,ll Y2,ll dt){
    upt(X1 , Y1 , dt ) ;
    upt(X2 + 1 , Y1 , -dt ) ;
    upt(X1 , Y2 + 1, -dt ) ;
    upt(X2 + 1, Y2 + 1, dt ) ; 
}
IL ll query(ll X1,ll Y1,ll X2,ll Y2){
    return
        calc(X2 , Y2) + calc(X1 - 1, Y1 - 1) -
        calc(X1 - 1 , Y2) - calc(X2 , Y1 - 1) ;
}

int main(){
    ll X1 , X2 , Y1 , Y2 , z; char c[3]; 
    scanf("X %lld %lld",&n,&m);
    while(scanf("%s",c)!=EOF){
        scanf("%lld%lld%lld%lld", &X1, &Y1, &X2, &Y2);
        if(c[0]=='L'){
            scanf("%lld", &z);
            add(X1 , Y1 , X2 , Y2 , z);
        }
        else printf("%lld\n",query(X1 , Y1 , X2 , Y2 ));
    }return 0;return 0;
    
}

转载于:https://www.cnblogs.com/Guess2/p/8459568.html

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