poj1195 Mobile phones 二维线段树和二维树状数组两种做法 树套树
Mobile phones
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 15849 | Accepted: 7321 |
Description
Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix.
Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.
Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.
Input
The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table.
The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.
Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30
The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.
Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30
Output
Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.
Sample Input
0 4 1 1 2 3 2 0 0 2 2 1 1 1 2 1 1 2 -1 2 1 1 2 3 3
Sample Output
3 4
操作0：S*S的矩阵置0 ，操作1：坐标(x,y)加v，操作2：[L,R]，[B,T]这个矩阵的和。
一开始没看清楚题，以为操作1也是在矩阵内加，准备写个带pushdown二维线段树。但是写pushdown的时候发现写不出来，感觉好像有一维要推的话必须把另一维整个推下去，这个能不能实现我不知道，反正我现在不会写。于是就没写带pushdown的。
其实二维线段树的思想就是一维线段树的点当成一维，想明白了并不难，就是代码写起来挺麻烦。
这题用二维树状数组做代码就短多了。树状数组c变成了二维，与一维类似，c[x][y]表示c[x-lowbit(x)][y-lowbit(y)]这个矩阵内的和，两重循环更新，sum(x,y)求[1,x],[1,y]这个矩阵的和，最后答案再用容斥原理减一下。
二维线段树
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXN=1030;
const int MAXNODE=MAXN*4;
const LL MOD=1000000007;
int N,M;
struct SegmentTree{
int sum[MAXNODE][MAXNODE];
void build_sub(int o,int L,int R,int id){
sum[id][o]=0;
if(L>=R) return;
int mid=L+(R-L)/2;
if(L<=mid) build_sub(o<<1,L,mid,id);
if(R>mid) build_sub(o<<1|1,mid+1,R,id);
}
void build(int o,int L,int R){
build_sub(1,1,M,o);
if(L>=R) return;
int mid=L+(R-L)/2;
if(L<=mid) build(o<<1,L,mid);
if(R>mid) build(o<<1|1,mid+1,R);
}
void update_sub(int o,int L,int R,int ql,int qr,int id,int v){
sum[id][o]+=v;
if(L>=R) return;
int mid=L+(R-L)/2;
if(ql<=mid) update_sub(o<<1,L,mid,ql,qr,id,v);
if(qr>mid) update_sub(o<<1|1,mid+1,R,ql,qr,id,v);
}
void update(int o,int L,int R,int ql,int qr,int subl,int subr,int v){
update_sub(1,1,M,subl,subr,o,v);
if(L>=R) return;
int mid=L+(R-L)/2;
if(ql<=mid) update(o<<1,L,mid,ql,qr,subl,subr,v);
if(qr>mid) update(o<<1|1,mid+1,R,ql,qr,subl,subr,v);
}
int query_sub(int o,int L,int R,int ql,int qr,int id){
if(ql<=L&&qr>=R) return sum[id][o];
int mid=L+(R-L)/2;
int ret=0;
if(ql<=mid) ret+=query_sub(o<<1,L,mid,ql,qr,id);
if(qr>mid) ret+=query_sub(o<<1|1,mid+1,R,ql,qr,id);
return ret;
}
int query(int o,int L,int R,int ql,int qr,int subl,int subr){
if(ql<=L&&qr>=R) return query_sub(1,1,M,subl,subr,o);
int mid=L+(R-L)/2;
int ret=0;
if(ql<=mid) ret+=query(o<<1,L,mid,ql,qr,subl,subr);
if(qr>mid) ret+=query(o<<1|1,mid+1,R,ql,qr,subl,subr);
return ret;
}
}tree;
int main(){
freopen("in.txt","r",stdin);
int op;
while(scanf("%d",&op)!=EOF){
if(op==0){
int s;
scanf("%d",&s);
N=M=s;
tree.build(1,1,N);
}
else if(op==1){
int x,y,a;
scanf("%d%d%d",&x,&y,&a);
x++;
y++;
tree.update(1,1,N,x,x,y,y,a);
}
else if(op==2){
int l,b,r,t;
scanf("%d%d%d%d",&l,&b,&r,&t);
l++;
b++;
r++;
t++;
printf("%d\n",tree.query(1,1,N,l,r,b,t));
}
else break;
}
return 0;
}
二维树状数组
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXN=1030;
const int MAXNODE=MAXN*4;
const LL MOD=1000000007;
int N;
int c[MAXN][MAXN];
int lowbit(int x){
return x&(-x);
}
void add(int x,int y,int v){
for(int i=x;i<=N;i+=lowbit(i))
for(int j=y;j<=N;j+=lowbit(j)) c[i][j]+=v;
}
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("in.txt","r",stdin);
int op;
while(scanf("%d",&op)!=EOF){
if(op==0){
int s;
scanf("%d",&s);
N=s;
memset(c,0,sizeof(c));
}
else if(op==1){
int x,y,a;
scanf("%d%d%d",&x,&y,&a);
x++;
y++;
add(x,y,a);
}
else if(op==2){
int l,b,r,t;
scanf("%d%d%d%d",&l,&b,&r,&t);
l++;
b++;
r++;
t++;
printf("%d\n",sum(r,t)-sum(l-1,t)-sum(r,b-1)+sum(l-1,b-1));
}
else break;
}
return 0;
}