Codeforces Round #137 (Div. 2)——B

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 "s_i x_i y_i", where s_i is one of the characters "с", "r" or "g", andx_i, y_i are two integers.

• If s_i = "c", then the current query is the query to swap columns with indexes x_i and y_i (1 ≤ x, y ≤ m, x ≠ y);
• If s_i = "r", then the current query is the query to swap rows with indexes x_i and y_i (1 ≤ x, y ≤ n, x ≠ y);
• If s_i = "g", then the current query is the query to obtain the number that located in the x_i-th row and in the y_i-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 (s_i = "g") print the required number. Print the answers to the queries in the order of the queries in the input.

Sample test(s)

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.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 1005int a[maxn][maxn];int r[maxn];//记录行变换int c[maxn];//记录列变换int main(){    int n,m,k;    scanf("%d%d%d",&n,&m,&k);    int i,j;    for(i=1; i<=n; i++)        for(j=1; j<=m; j++)            scanf("%d",&a[i][j]);    //存储数组初始化    for(i=1;i<=n;i++) r[i]=i;    for(j=1;j<=m;j++) c[j]=j;    while(k--)    {        getchar();        char ch;        int x,y;        scanf("%c%d%d",&ch,&x,&y);        if(ch=='g') printf("%d\n",a[r[x]][c[y]]);        if(ch=='c')        {                //交换第x,y列                int t=c[x];                c[x]=c[y];                c[y]=t;        }        if(ch=='r')        {                //交换第x,y行                int t=r[x];                r[x]=r[y];                r[y]=t;        }    }    return 0;}