【Codeforces】-222B-Cosmic Tables

B. Cosmic Tables

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Free Meteor Association (FMA) has got a problem: as meteors are moving, the Universal Cosmic Descriptive Humorous Program (UCDHP) needs to add a special module that would analyze this movement.

UCDHP stores some secret information about meteors as an n × m table with integers in its cells. The order of meteors in the Universe is changing. That's why the main UCDHP module receives the following queries:

• The query to swap two table rows;
• The query to swap two table columns;
• The query to obtain a secret number in a particular table cell.

As the main UCDHP module is critical, writing the functional of working with the table has been commissioned to you.

Input

The first line contains three space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 500000) — the number of table columns and rows and the number of queries, correspondingly.

Next n lines contain m space-separated numbers each — the initial state of the table. Each number p in the table is an integer and satisfies the inequality 0 ≤ p ≤ 106.

Next k lines contain queries in the format "si xi yi", where si is one of the characters "с", "r" or "g", and xi, yi are two integers.

• If si = "c", then the current query is the query to swap columns with indexes xi and yi (1 ≤ x, y ≤ m, x ≠ y);
• If si = "r", then the current query is the query to swap rows with indexes xi and yi (1 ≤ x, y ≤ n, x ≠ y);
• If si = "g", then the current query is the query to obtain the number that located in the xi-th row and in the yi-th column (1 ≤ x ≤ n, 1 ≤ y ≤ m).

The table rows are considered to be indexed from top to bottom from 1 to n, and the table columns — from left to right from 1 to m.

Output

For each query to obtain a number (si = "g") print the required number. Print the answers to the queries in the order of the queries in the input.

Examples

input

3 3 51 2 34 5 67 8 9g 3 2r 3 2c 2 3g 2 2g 3 2

output

896

input

2 3 31 2 43 1 5c 2 1r 1 2g 1 3

output

5

Note

Let's see how the table changes in the second test case.

After the first operation is fulfilled, the table looks like that:

2 1 4

1 3 5

After the second operation is fulfilled, the table looks like that:

1 3 5

2 1 4

So the answer to the third query (the number located in the first row and in the third column) will be 5.

暴力超时！！！用两个一位数组单独直接表示行和列，交换行或者列直接交换数组里的下标即可。

#include<cstdio>#include<algorithm>using namespace std; int d[1010][1010];int c[1010],r[1010];int n,m,k;void lll(){int l;l=max(n,m);for(int i=1;i<=l;i++)c[i]=r[i]=i;}int main(){scanf("%d %d %d",&n,&m,&k);for(int i=1;i<=n;i++){for(int j=1;j<=m;j++)scanf("%d",&d[i][j]);}lll();while(k--){char e[4];int a,b;scanf("%s %d %d",e,&a,&b);if(e[0]=='g')printf("%d\n",d[r[a]][c[b]]);else if(e[0]=='c')//交换列 {swap(c[a],c[b]); }else if(e[0]=='r')//交换行 {swap(r[a],r[b]);}}return 0;}