CodeForces 222B Cosmic Tables

Cosmic Tables

Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 222B

Description

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 rows, columns 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", and x_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 Input

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

Hint

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<stdio.h>
#define max1 1005
using namespace std;
int a[max1][max1],r[max1],c[max1];
int main()
{
    int n,m,t;
    scanf("%d%d%d",&n,&m,&t);
    for(int i=;i<=n;i++)
        r[i]=i;
    for(int i=;i<=m;i++)
        c[i]=i;
    for(int i=;i<=n;i++)
    {
        for(int j=;j<=m;j++)
        {
            scanf("%d",&a[i][j]);
        }
    }
    getchar();
    char ch;
    for(int i=;i<t;i++)
    {
        //getchar();
        scanf("%c",&ch);
        int n1,n2;
        scanf("%d%d",&n1,&n2);getchar();
        if(ch=='r')
        {

            int temp=r[n1];
            r[n1]=r[n2];
            r[n2]=temp;
        }
        else if(ch=='c')
        {
            int temp=c[n1];
            c[n1]=c[n2];
            c[n2]=temp;
        }
        else if(ch=='g')
        {
            printf("%d\n",a[r[n1]][c[n2]]);
        }
    }
    return ;
}

模拟，根据题目操作一遍就行。注意要用scanf，cin会超时。