【CodeForces】222B - Cosmic Tables（思维）

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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 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.

以前做过一个，题意是把给定的行（列）加到另一行（列）上，用的是并查集，感觉那个思路很好。这道题要求是交换，就不能用并查集了。

刚开始绕弯很多，主要是没搞懂数组的值表示的含义。

下面代码，r [ ] 、c [ ] 数组分别表示现在的行和列表示原有的哪一行（列）。（这点一定要清楚，要不写代码的时候就搞晕了）

那么明白了含义后，进行交换操作就简单多了，（媳妇儿说坑坑坑），把当前行（列）表示的真正数值直接交换就行了。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int h,w,k;
int mapp[1011][1011];
int c[1011],r[1011];		//c为列，r为行: 表示第几行（列）的数表示原来的第几行（列） 
void init()
{
	int l;
	l = max (w,h);
	for (int i = 1 ; i <= l ; i++)
		c[i] = r[i] = i;
}
int main()
{
	char op[4];
	int x,y;
	while (~scanf ("%d %d %d",&h,&w,&k))
	{
		init();		//初始化 
		for (int i = 1 ; i <= h ; i++)
			for (int j = 1 ; j <= w ; j++)
				scanf ("%d",&mapp[i][j]);
		for (int i = 1 ; i <= k ; i++)
		{
			scanf ("%s %d %d",op,&x,&y);
			if (op[0] == 'r')
				swap (r[x] , r[y]);
			else if (op[0] == 'c')
				swap (c[x] , c[y]);
			else
				printf ("%d\n",mapp[r[x]][c[y]]);
		}
	}
	return 0;
}