UFOs

时间：2016-08-24 12:38:26 阅读：204 评论：0 收藏：0 [点我收藏+]

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UFOs

Time limit: 2.0 second
Memory limit: 64 MB

Vasya is a ufologist and his duties include observing Unidentified Flying Objects (UFOs) in the part of space bounded by a cube N × N × N. The cube is divided into cubic sectors 1 × 1 × 1. During the observation, the following events may happen:

several new UFOs emerge in a certain sector;
several UFOs disappear in a certain sector;
Vasya‘s boss may ask him how many UFOs there are in a part of space consisting of several sectors.

At the moment when Vasya starts his observations there are no UFOs in the whole space.

Input

The first line contains an integer N (1 ≤ N ≤ 128). The coordinates of sectors are integers from 0 to N–1.

Then there are entries describing events, one entry per line. Each entry starts with a number M.

If M is 1, then this number is followed by four integers x (0 ≤ x < N), y (0 ≤ y < N), z (0 ≤ z < N), K (–20000 ≤ K ≤ 20000), which are coordinates of a sector and the change in the number of UFOs in this sector. The number of UFOs in a sector cannot become negative.
If M is 2, then this number is followed by six integers x1, y1, z1, x2, y2, z2 (0 ≤ x1 ≤ x2 <N, 0 ≤ y1 ≤ y2 < N, 0 ≤ z1 ≤ z2 < N), which mean that Vasya must compute the total number of UFOs in sectors (x, y, z) belonging to the volume: x1 ≤ x ≤ x2, y1 ≤ y ≤ y2, z1 ≤ z ≤z2.
If M is 3, it means that Vasya is tired and goes to sleep. This entry is always the last one.

The number of entries does not exceed 100002.

Output

For each query, output in a separate line the required number of UFOs.

Sample

input	output
2 2 1 1 1 1 1 1 1 0 0 0 1 1 0 1 0 3 2 0 0 0 0 0 0 2 0 0 0 0 1 0 1 0 1 0 -2 2 0 0 0 1 1 1 3	0 1 4 2

input

output

2
2 1 1 1 1 1 1
1 0 0 0 1
1 0 1 0 3
2 0 0 0 0 0 0
2 0 0 0 0 1 0
1 0 1 0 -2
2 0 0 0 1 1 1
3

分析：三维树状数组，求和时类似于容斥；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t;
ll a[129][129][129];
void add(int x,int y,int z,int w)
{
    for(int i=x;i<=n;i+=(i&(-i)))
        for(int j=y;j<=n;j+=(j&(-j)))
            for(int k=z;k<=n;k+=(k&(-k)))
                a[i][j][k]+=w;
}
ll get(int x,int y,int z)
{
    ll ans=0;
    for(int i=x;i;i-=(i&(-i)))
        for(int j=y;j;j-=(j&(-j)))
            for(int k=z;k;k-=(k&(-k)))
                ans+=a[i][j][k];
    return ans;
}
int main()
{
    int i,j;
    scanf("%d",&n);
    while(~scanf("%d",&m)&&m!=3)
    {
        if(m==1)
        {
            int b[4];
            rep(i,0,3)scanf("%d",&b[i]);
            add(++b[0],++b[1],++b[2],b[3]);
        }
        else
        {
            int b[6];
            rep(i,0,5)scanf("%d",&b[i]),b[i]++;
            printf("%lld\n",get(b[3],b[4],b[5])
                   -get(b[0]-1,b[4],b[5])-get(b[3],b[1]-1,b[5])-get(b[3],b[4],b[2]-1)
                   +get(b[0]-1,b[1]-1,b[5])+get(b[3],b[1]-1,b[2]-1)+get(b[0]-1,b[4],b[2]-1)
                   -get(b[0]-1,b[1]-1,b[2]-1));
        }
    }
    //system("Pause");
    return 0;
}

UFOs

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原文地址：http://www.cnblogs.com/dyzll/p/5802139.html

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