BZOJ 1176([Balkan2007]Mokia-CDQ分治-分治询问)

1176: [Balkan2007]Mokia

Time Limit: 30 Sec Memory Limit: 162 MB
Submit: 185 Solved: 94
[ Submit][ Status]

Description

维护一个W*W的矩阵，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

Output

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

Input file Output file Meaning
0 4 Table size is , filled with zeroes.

1 2 3 3 Add 3 customers at (2, 3).
2 1 1 3 3 Query sum of rectangle , .

3 Answer.
1 2 2 2 Add 2 customers at (2, 2).
2 2 2 3 4 Query sum of rectangle , .

5 Answer
3 Exit your program.

裸CDQ...1A

注意几个关键部分的写法

步骤：

1.读入询问，按x排序

2.将[L,R]中的数分为前部分操作，后部分操作（各部分仍保持X升序）

3.将前面对后面的影响记录ans

4.复原影响

5.递归[L,M],[M+1,R]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
#include<cassert>
#include<climits>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a))
#define MEMI(a) memset(a,127,sizeof(a))
#define MEMi(a) memset(a,128,sizeof(a))
#define INF (2139062143)
#define F (1000000009)
#define MAXM (640000+10)
#define MAXQ (10000+10)
#define MAXN (2000000+10)
typedef long long ll;
int n,m,q;
ll ans[MAXQ];
struct comm
{int no,x,y,v,q,type;comm():no(0),x(0),y(0),v(0),q(0),type(0){}comm(int _no,int _x,int _y,int _v,int _q,int _type):no(_no),x(_x),y(_y),v(_v),q(_q),type(_type){}friend bool operator<(comm a,comm b){return a.x<b.x;}
}ask[MAXM];
struct arr_tree
{ll a[MAXN];arr_tree(){memset(a,0,sizeof(a));}void add(int x,ll c){for(int i=x;i<=n;i+=i&(-i)) a[i]+=c;}int qur(int x){ll ans=0;for(int i=x;i;i-=i&(-i)) ans+=a[i];return ans;   }void clear(){memset(a,0,sizeof(a));}
}T;
comm tmp[MAXM];
void solve(int l,int r)
{if (l==r) return;int m=l+r>>1;int s1=l-1,s2=m;Fork(i,l,r) tmp[ask[i].no<=m?++s1:++s2]= ask[i];memcpy(ask+l,tmp+l,sizeof(comm)*(r-l+1));int now=l;Fork(i,m+1,r) //遍历询问 {if (ask[i].type==2){while (ask[now].x<=ask[i].x&&now<=m){if (ask[now].type==1) T.add(ask[now].y,ask[now].v);now++;}ans[ask[i].q]+=ask[i].v*T.qur(ask[i].y);         }}now--;while (l<=now) {if (ask[now].type==1) T.add(ask[now].y,-ask[now].v);now--;}solve(l,m),solve(m+1,r);
}
bool work()
{scanf("%d",&n);int type,no=0,x,y,x1,y1,x2,y2,v,q=0;memset(ans,0,sizeof(ans));T.clear();while (scanf("%d",&type)){if (type==0||type==3) break;else if (type==1){scanf("%d%d%d",&x,&y,&v);no++;ask[no]=comm(no,x,y,v,0,1);}else if (type==2){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);q++;no++;ask[no]=comm(no,x1-1,y1-1,1,q,2);no++;ask[no]=comm(no,x2,y2,1,q,2);no++;ask[no]=comm(no,x1-1,y2,-1,q,2);no++;ask[no]=comm(no,x2,y1-1,-1,q,2);         }}sort(ask+1,ask+1+no);solve(1,no);For(i,q) cout<<ans[i]<<endl;return type==0;
}
int main()
{//freopen("bzoj1176.in","r",stdin);int type;scanf("%d",&type);if (type==3) return 0;while (work());//cout<<"END"<<endl;while (1);return 0;
}

转载于:https://www.cnblogs.com/dyllove98/p/3186980.html