【BOI2007】【BZOJ1176】Mokia

1176: [Balkan2007]Mokia

Time Limit: 30 Sec Memory Limit: 162 MB

Submit: 1059 Solved: 432

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Description

维护一个W*W的矩阵，初始值均为S.每次操作可以增加某格子的权值,或询问某子矩阵的总权值.修改操作数M<=160000,询问数Q<=10000,W<=2000000.

Input

第一行两个整数,S,W;其中S为矩阵初始值;W为矩阵大小

接下来每行为一下三种输入之一(不包含引号):

“1 x y a”

“2 x1 y1 x2 y2”

“3”

输入1:你需要把(x,y)(第x行第y列)的格子权值增加a

输入2:你需要求出以左上角为(x1,y1),右下角为(x2,y2)的矩阵内所有格子的权值和,并输出

输入3:表示输入结束

Output

对于每个输入2,输出一行,即输入2的答案

Sample Input

0 4

1 2 3 3

2 1 1 3 3

1 2 2 2

2 2 2 3 4

3

Sample Output

3

5

HINT

保证答案不会超过int范围

Source

cdq分治的模板题然而我现在才做这个题。。。

差分一下询问操作。对所有操作按y排序。

树状数组维护一下。然后就可以了。

自从看过了Tsinsen上的姿势分

我写代码都开始丧心病狂了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 200000
#define SIZE 2000010
#define lowbit(x) (x&(-x))
using namespace std;
int w;
int top,opt,L,R,l,r,delta,Top;
struct Query
{
    int op;
    int x,y,A;
    int t,id;
    bool operator <(const Query& a)const
    {
        if (x == a.x && y == a.y) return op < a.op;
        if (x == a.x) return y < a.y;
        return x < a.x;
    }
}que[MAXN],newq[MAXN];
int ans[MAXN],c[SIZE];
inline void in(int &x)
{
    x=0;char ch = getchar();
    while (!(ch >= ‘0‘ && ch <= ‘9‘))   ch = getchar();
    while (ch >= ‘0‘ && ch <= ‘9‘)  x = x * 10 + ch - ‘0‘,ch = getchar();
}
inline void add(int i,int x)
{
    while (i && i <= w) c[i] += x,i += lowbit(i);
}
inline int query(int i)
{
    int ret = 0;
    while (i) ret += c[i],i -= lowbit(i);
    return ret;
}
inline void Solve(int l,int r)
{
    int mid = (l + r) >> 1,tp1 = l,tp2 = mid + 1;
    if (l == r) return;
    for (int i = l;i <= r;i++)
    {
        if (que[i].t <= mid && que[i].op == 1)  add(que[i].y,que[i].A);
        if (que[i].t > mid && que[i].op == 2)   ans[que[i].id] += query(que[i].y) * que[i].A;
    }
    for (int i = l;i <= r;i++)
        if (que[i].t <= mid && que[i].op == 1) add(que[i].y,-que[i].A);
    for (int i = l;i <= r;i++)
        if (que[i].t <= mid) newq[tp1++] = que[i];
        else newq[tp2++] = que[i];
    memcpy(que+l,newq+l,sizeof(Query)*(r - l + 1));
    Solve(l,mid);Solve(mid+1,r);
}
int main()
{
    freopen("mokia.in","r",stdin);
    freopen("mokia.out","w",stdout);
    in(opt);in(w);
    while (1)
    {
        in(opt);
        if (opt == 3) break;
        switch (opt)
        {
            case 1:
                in(L);in(R);in(delta);
                que[++top].op = opt;que[top].x = L;que[top].y = R;que[top].A = delta;que[top].t = top;
                break;
            case 2:
                in(L);in(R);in(l);in(r);
                que[++top].op = opt;que[top].x = L - 1;que[top].y = R - 1;que[top].t = top;que[top].A = 1;que[top].id = ++Top;
                que[++top].op = opt;que[top].x = L - 1;que[top].y = r;que[top].t = top;que[top].A = -1;que[top].id = Top;
                que[++top].op = opt;que[top].x = l;que[top].y = R - 1;que[top].t = top;que[top].A = -1;que[top].id = Top;
                que[++top].op = opt;que[top].x = l;que[top].y = r;que[top].t = top;que[top].A = 1;que[top].id = Top;
                break;
        }
    }
    sort(que + 1,que + top + 1);
    Solve(1,top);
    for (int i = 1;i <= Top;i++)    printf("%d\n",ans[i]);
}