CodeForces 222B Cosmic Tables

时间：2023-01-21 06:00:48
Cosmic Tables

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

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 "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 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。

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