POJ 2155——Matrix（树套树，二维树状数组，二维线段树）

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 18460 Accepted: 6950

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

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题目大意：

有一个n*n的矩形平面，里面每个点初始为0，有两种操作：

1.将左上角为（x1，y1），右下角为（x2，y2）的矩形区域中的每个点，把0变成1，把1变成0

2.查询点（x，y）当前的值

思路：

对于翻转矩形区域，一种是向上，一种向下，跟一维是类似的，辅助数组c[][]来记录修改次数

二维树状数组：

向上修改，向下统计：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int n,m;int c[1010][1010];void update(int x,int y){    for(int i=x;i<=n;i+=i&-i){        for(int j=y;j<=n;j+=j&-j){            c[i][j]++;        }    }}int getsum(int x,int y){    int sum=0;    for(int i=x;i>0;i-=i&-i){        for(int j=y;j>0;j-=j&-j){            sum+=c[i][j];        }    }    return sum;}int main(){    int T;    cin>>T;    while(T--){        memset(c,0,sizeof(c));        scanf("%d %d",&n,&m);        getchar();        while(m--){            char ch;            ch=getchar();            if(ch=='C'){                int x1,y1,x2,y2;                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);                getchar();                update(x1,y1);                update(x2+1,y2+1);                update(x2+1,y1);                update(x1,y2+1);            }            if(ch=='Q'){                int x,y;                scanf("%d %d",&x,&y);                getchar();                printf("%d\n",getsum(x,y)&1);            }        }        if(T) printf("\n");    }    return 0;}

向下修改，向上统计：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int n,m;int c[1010][1010];void update(int x,int y){    for(int i=x;i>0;i-=i&-i){        for(int j=y;j>0;j-=j&-j){            c[i][j]++;        }    }}int getsum(int x,int y){    int sum=0;    for(int i=x;i<=n;i+=i&-i){        for(int j=y;j<=n;j+=j&-j){            sum+=c[i][j];        }    }    return sum;}int main(){    int T;    cin>>T;    while(T--){        memset(c,0,sizeof(c));        scanf("%d %d",&n,&m);        getchar();        while(m--){            char ch;            ch=getchar();            if(ch=='C'){                int x1,y1,x2,y2;                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);                getchar();                update(x2,y2);                update(x1-1,y2);                update(x2,y1-1);                update(x1-1,y1-1);            }            if(ch=='Q'){                int x,y;                scanf("%d %d",&x,&y);                getchar();                printf("%d\n",getsum(x,y)&1);            }        }        if(T) printf("\n");    }    return 0;}

二维线段树：

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define rep(i,n) for(int i=0;i<n;++i)#define mes(s,c) memset(s,c,sizeof(s))const int maxn=1001;using namespace std;int sum[maxn<<2][maxn<<2];int n,m;int ans;void Subbuild(int st,int l,int r,int rt){    sum[st][rt]=0;    if(l==r) return ;    int m=(l+r)>>1;    Subbuild(st,lson);    Subbuild(st,rson);}void Build(int l,int r,int rt){    Subbuild(rt,1,n,1);    if(l==r) return;    int m=(l+r)>>1;    Build(lson);    Build(rson);}void Subupdate(int L,int R,int st,int l,int r,int rt){    if(L<=l&&r<=R){        sum[st][rt]^=1;        return ;    }    int m=(l+r)>>1;    if(L<=m) Subupdate(L,R,st,lson);    if(m<R) Subupdate(L,R,st,rson);}void Update(int x1,int y1,int x2,int y2,int l,int r,int rt){    if(x1<=l&&r<=x2){        Subupdate(y1,y2,rt,1,n,1);        return ;    }    int m=(l+r)>>1;    if(x1<=m) Update(x1,y1,x2,y2,lson);    if(m<x2) Update(x1,y1,x2,y2,rson);}void Subquery(int L,int R,int st,int l,int r,int rt){    ans^=sum[st][rt];    if(l==r) return;    int m=(l+r)>>1;    if(R<=m) Subquery(L,R,st,lson);    else Subquery(L,R,st,rson);}void Query(int L,int R,int l,int r,int rt){    Subquery(L,R,rt,1,n,1);    if(l==r) return ;    int m=(l+r)>>1;    if(L<=m) Query(L,R,lson);    else Query(L,R,rson);}int main(){    int T;    cin>>T;    while(T--){        scanf("%d%d",&n,&m);        Build(1,n,1);        char op[10];        int x1,y1,x2,y2,x,y;        while(m--){            scanf("%s",op);            if(op[0]=='C'){                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                Update(x1,y1,x2,y2,1,n,1);            }else{                ans=0;                scanf("%d%d",&x,&y);                Query(x,y,1,n,1);                printf("%d\n",ans);            }        }        if(T) puts(" ");    }    return 0;}