【CODEFORCES】 B. 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.

Sample test(s)

input

3 3 5
1 2 3
4 5 6
7 8 9
g 3 2
r 3 2
c 2 3
g 2 2
g 3 2

output

8
9
6

input

2 3 3
1 2 4
3 1 5
c 2 1
r 1 2
g 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.

题解：这一题很巧妙，用b[i]表示现在的第i行是原来的第几行，c[i]表示现在的第i列是原来的第几列。然后只要交换b[x],b[y]就可以了，答案即是a[b[x]][c[y]]。

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int n,k,i,t,a[100005];
int main()
{
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    i=k;
    while (a[i]==a[i-1] && i>=0) i--;
    t=i;
    for (int i=k;i<=n;i++) if (a[i]!=a[k])
    {
        cout <<-1<<endl;
        return 0;
    }
    cout <<t-1<<endl;
    return 0;
}