【树转数组】poj1195

/*
二维的树状数组:
更新某个元素时:
NO.1:c[n1],c[n2],c[n3],....,c[nm];
其中n1 = i,n(i+1) = ni+lowbit(ni);
nm+lowbit(nm)的值应该大于元素个数N。
NO.2:sum(k)=c[n1]+c[n2]+...+c[nm];
其中nm=k,n(i-1)=ni-lowbit(ni);
n1-lowbit(n1)的值应该小于0
----------------------------------------------------------------------
用二维的c[i][j]存储长条的和。。。、
---------------------------------------
元素的值加一个数:
void update(int x, int y, int a)
{
    for(int i=x; i<=s; i+=lowbit(i))
    {
        for(int j=y; j<=s; j+=lowbit(j))
            c[i][j] += a;
    }
}
---------------------------------------------------
int sum(int x, int y)矩阵0到x和0到y的元素和
{
    int ret = 0;
    for(int i=x; i>0; i-=lowbit(i))
    {
        for(int j=y; j>0; j-=lowbit(j))
            ret += c[i][j];
    }
    return ret;
}
------------------------------------------------
*/
#include 
#include 
#include 
#define INF 0x3f3f3f3f

using namespace std;

int op,s;
int x,y,a;
int l,b,r,t;
int c[1050][1050];

int lowbit(int x)
{
    return x & (-x);
}

void update(int x, int y, int a)
{
    for(int i=x; i<=s; i+=lowbit(i))
    {
        for(int j=y; j<=s; j+=lowbit(j))
            c[i][j] += a;
    }
}

int sum(int x, int y)
{
    int ret = 0;
    for(int i=x; i>0; i-=lowbit(i))
    {
        for(int j=y; j>0; j-=lowbit(j))
            ret += c[i][j];
    }
    return ret;
}

int main()
{
    //freopen("input.txt","r",stdin);
    while(1)
    {
        scanf("%d",&op);
        if(op == 0)
        {
            scanf("%d",&s);
            memset(c, 0, sizeof(c));
        }
        if(op == 1)
        {
            scanf("%d%d%d",&x,&y,&a);
            update(x+1, y+1, a);
        }
        if(op == 2)
        {
            scanf("%d%d%d%d",&l, &b,&r,&t);
            l++;
            b++;
            r++;
            t++;
            printf("%d\n",sum(r, t) - sum(r, b-1) - sum(l-1, t) + sum(l-1, b-1) );
        }
        if(op == 3)
            return 0;
    }
}

----------------------------------

快。。。。。

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