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HDU 2155 Matrix

2024-07-26 07:14:03   218人阅读

Matrix

Time Limit: 3000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2155
64-bit integer IO format: %lld      Java class name: Main
 
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

Source

POJ Monthly,Lou Tiancheng
 
解题：二维线段树，第一次玩这鬼玩意，调试了N久。。。挫。。。。
 
 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 1001;18 struct subtree{19     int lt,rt;20     bool val;21 };22 struct node{23     int lt,rt;24     subtree sb[maxn<<2];25 };26 node tree[maxn<<2];27 int n,m,ans;28 void sub(int lt,int rt,int v,subtree *sb){29     sb[v].lt = lt;30     sb[v].rt = rt;31     sb[v].val = false;32     if(lt == rt) return;33     int mid = (lt + rt)>>1;34     sub(lt,mid,v<<1,sb);35     sub(mid+1,rt,v<<1|1,sb);36 }37 void build(int lt,int rt,int v){38     tree[v].lt = lt;39     tree[v].rt = rt;40     sub(1,n,1,tree[v].sb);41     if(lt == rt) return;42     int mid = (lt + rt)>>1;43     build(lt,mid,v<<1);44     build(mid+1,rt,v<<1|1);45 }46 void subupdate(int lt,int rt,int v,subtree *sb){47     if(sb[v].lt >= lt && sb[v].rt <= rt){48         sb[v].val ^= 1;49         return;50     }51     if(lt <= tree[v<<1].rt) subupdate(lt,rt,v<<1,sb);52     if(rt >= tree[v<<1|1].lt) subupdate(lt,rt,v<<1|1,sb);53 }54 void update(int lt,int rt,int y1,int y2,int v){55     if(tree[v].lt >= lt && tree[v].rt <= rt){56         subupdate(y1,y2,1,tree[v].sb);57         return;58     }59     if(lt <= tree[v<<1].rt) update(lt,rt,y1,y2,v<<1);60     if(rt >= tree[v<<1|1].lt) update(lt,rt,y1,y2,v<<1|1);61 62 }63 void subquery(int y,int v,subtree *sb){64     ans ^= sb[v].val;65     if(sb[v].lt == sb[v].rt) return;66     if(y <= sb[v<<1].rt) subquery(y,v<<1,sb);67     else subquery(y,v<<1|1,sb);68 }69 void query(int x,int y,int v){70     subquery(y,1,tree[v].sb);71     if(tree[v].lt == tree[v].rt) return;72     if(x <= tree[v<<1].rt) query(x,y,v<<1);73     else query(x,y,v<<1|1);74 }75 int main(){76     int t,x1,y1,x2,y2;77     char s[5];78     scanf("%d",&t);79     while(t--){80         scanf("%d %d",&n,&m);81         build(1,n,1);82         for(int i = 0; i < m; ++i){83             scanf("%s",s);84             if(s[0] == ‘Q‘){85                 ans = 0;86                 scanf("%d %d",&x1,&y1);87                 query(x1,y1,1);88                 printf("%d\n",ans);89             }else if(s[0] == ‘C‘){90                 scanf("%d %d %d %d",&x1,&y1,&x2,&y2);91                 update(x1,x2,y1,y2,1);92             }93         }94         puts("");95     }96     return 0;97 }
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HDU 2155 Matrix


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