Bestcoders

Senior's Fish

Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 242    Accepted Submission(s): 50

Problem Description

Xuejiejie loves to eat fish. One day, she goes to a pond for fishing. The pond can be seen as a two-dimensional surface, and her fishing net can be seen as a rectangle. One of the edges of the rectangle is parallel to the x-axis ,and another is parallel to y-axis. The fishes can be seen as points. Sometimes the fishes will enter the net, and sometimes they will leave the net. So, Xuejiejie doesn't know the appropriate time to draw the net in when she will get as many fishes as possible. 

Xuejiejie assigns each fish with a number, from 1 to  n ,  n  being the total number. There are 2 types of movements of the fishes:

1 l r d : the fishes index between the interval [l，r] moved towards the x-axis for length d (For example , if a fish's current position is  (x,y)  , after moving , its position will change to  (x+d,y) . )

2 l r d : the fishes index between the interval [l，r] moved towards the y-axis for length d (For example , if a fish's current position is  (x,y) , after moving , its position will change to  (x,y+d) . )

And sometimes Xuejiejie will ask you some questions.

 

Input

In the first line there is an integer  T , indicates the number of test cases.

In each case:

The first line includes an integer  n  indicating the total number of fish.

The second line includes 4 integers  x1 ， y1 ， x2 ， y2 ，indicating the position of the fishing net.  (x1,y1)  means the lower-left position, and  (x2,y2)  means the top-right position.

The next n lines: each line includes  x[i],y[i] ; means the fish i's initial position.

The next line includes an integer  m .

The next  m  lines describe the events you have to deal with.

In each line the first integer is  c  ( 1≤c≤3 ), which indicates the type of events.

1 l r d : the fish index between the interval [l，r] moved towards the x-axis for length d

2 l r d : the fish index between the interval [l，r] moved towards the y-axis for length d

3 l r : query the number of the fish index between the interval [l，r] which are in the net(including the one in the border)

1≤n,m≤100000 ,  1≤l≤r≤n .  1≤d≤109 ,  x1≤x2 ,  y1≤y2 。 Ensure that any time all involved coordinate values in the range of [−109,109] 。

 

Output

In each case:

For each type 3 events, output a integer which means the answer.

 

Sample Input

   
   
   
   

    
    
    
    1
5
1 1 5 5
1 1
2 2
3 3
4 4
5 5
3
3 1 5
1 2 4 2
3 1 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5
4
   
   
   
   

 

Source

BestCoder Round #47 ($)

 

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题解：用线段树维护每个点左下角内鱼的数目，因为渔网有四条边，根据限制条件的不同，需要维护四颗线段树，其限制条件分别不能从x or y 轴越过，否则该鱼不参与计数，设ans[0 ~ 3]分别为渔网右上角、左上角、右下角、左下角四个点左下方鱼的数目，则答案为: Ans = ans[0] - ans[1] - ans[2] + ans[3].

代码 I : (第一个写的，为超时版本， 正在努力修改)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
#define INF 0x7ffffff
#define eps 1e-8
#define maxn 100000 + 10
#define lson L, mid, rt<<1
#define rson mid+1, R, rt<<1|1

int X[4], Y[4], x[2], y[2];
int n, m;
int xx[maxn], yy[maxn];

struct Node
{
    int sum, mx, my, addx, addy; ///addx addy 相当于lazy标记
}T[4][maxn<<2];

void Pushup(int k, int rt)
{
    int l = rt<<1, r = rt<<1|1;
    T[k][rt].sum = T[k][l].sum + T[k][r].sum;
    T[k][rt].mx = max(T[k][l].mx, T[k][r].mx);
    T[k][rt].my = max(T[k][l].my, T[k][r].my);
}

void Pushdown(int k, int rt)
{
    int l = rt<<1, r = rt<<1|1;
    int t1 = T[k][rt].addx, t2 = T[k][rt].addy;

    T[k][l].addx += t1, T[k][r].addx += t1;
    T[k][l].addy += t2, T[k][r].addy += t2;
    T[k][l].mx += t1,   T[k][r].mx += t1;
    T[k][l].my += t2,   T[k][r].my += t2;

    T[k][rt].addx = T[k][rt].addy = 0;
}

void Bulid(int k, int L, int R, int rt) ///递归构造线段树
{
    T[k][rt].addx = T[k][rt].addy = 0;
    if(L == R)
    {
        if(xx[L]<=X[k] && yy[L]<=Y[k])
        {
            T[k][rt].sum = 1;
            T[k][rt].mx = xx[L], T[k][rt].my = yy[L];
        }
        else
        {
            T[k][rt].sum = 0;
            T[k][rt].mx = T[k][rt].my = -INF;
        }
        return ;
    }
    int mid = (L+R)>>1;
    Bulid(k, lson);
    Bulid(k, rson);
    Pushup(k, rt);
}

void Update(int k, int flag, int l, int r, int d, int L, int R, int rt)
{
    if(l<=L && r>=R)
    {
        if(flag) /// 1 -- x
        {
            T[k][rt].addx += d, T[k][rt].mx += d;
            if(T[k][rt].mx <= X[k])
                return ;
        }
        else    ///  0 -- y
        {
            T[k][rt].addy += d, T[k][rt].my += d;
            if(T[k][rt].my <= Y[k])
                return ;
        }
        if(L == R)
        {
            T[k][rt].sum = 0;
            T[k][rt].mx = T[k][rt].my = -INF;
            return ;
        }
    }
    int mid = (L+R)>>1;
    if(T[k][rt].addx || T[k][rt].addy) Pushdown(k, rt);
    if(r <= mid) Update(k, flag, l, r, d, lson);
    else if(l > mid) Update(k, flag, l, r, d, rson);
    else
    {
        Update(k, flag, l, mid, d, lson);
        Update(k, flag, mid+1, r, d, rson);
    }
    Pushup(k, rt);
}

int Query(int k, int l, int r, int L, int R, int rt)
{
    if(l<=L && r>=R)
        return T[k][rt].sum;
    int mid = (L+R)>>1;
    if(T[k][rt].addx || T[k][rt].addy) Pushdown(k, rt);
    if(r <= mid) Query(k, l, r, lson);
    else if(l > mid) Query(k, l, r, rson);
    else
      return Query(k, l, mid, lson) + Query(k, mid+1, r, rson);
}

int ans[4];

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);///
        scanf("%d%d%d%d", &x[0], &y[0], &x[1], &y[1]);
        X[0] = x[1], Y[0] = y[1];
        X[1] = x[0]-1, Y[1] = y[1];
        X[2] = x[1], Y[2] = y[0]-1;
        X[3] = x[0]-1, Y[3] = y[0]-1;
        for(int i=1; i<=n; i++)
            scanf("%d%d", &xx[i], &yy[i]);///

        for(int i=0; i<4; i++)
            Bulid(i, 1, n, 1);

        scanf("%d", &m);///
        for(int i=0; i<m; i++)
        {
            int t, l, r, d;
            scanf("%d%d%d", &t, &l, &r);
            if(t < 3) scanf("%d", &d);
            if(t == 1)
            {
                for(int i=0; i<4; i++)
                    Update(i, 1, l, r, d, 1, n, 1);
            }
            else if(t == 2)
            {
                for(int i=0; i<4; i++)
                    Update(i, 0, l, r, d, 1, n, 1);
            }
            else
            {
                for(int i=0; i<4; i++)
                    ans[i] = Query(i, l, r, 1, n, 1);
                printf("%d\n", ans[0] - ans[1] - ans[2] + ans[3]);
            }
        }
    }
    return 0;
}

/*

2

5
1 1 5 5
1 1
2 2
3 3
4 4
5 5
3
3 1 5
1 2 4 2
3 1 5

3
1 1 5 5
1 5
2 5
3 5
4
2 2 5 1
3 1 1
3 1 2
3 1 5

*/