Codeforces Round #147 (Div. 2) / 237B Young Table (搜索)

B. Young Table
http://codeforces.com/problemset/problem/237%2FB

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

You've got table a, consisting of n rows, numbered
from 1 to n. The i-th line of table a contains ci cells,
at that for all i (1 < i ≤ n) holdsci ≤ ci - 1.

Let's denote s as the total number of cells of table a,
that is,

.
We know that each cell of the table contains a single integer from 1 to s,
at that all written integers are distinct.

Let's assume that the cells of the i-th row of table a are
numbered from 1 to ci,
then let's denote the number written in the j-th cell of thei-th
row as ai, j.
Your task is to perform several swap operations to rearrange the numbers in the table so as to fulfill the following conditions:

for all i, j (1 < i ≤ n; 1 ≤ j ≤ ci) holds ai, j > ai - 1, j;

for all i, j (1 ≤ i ≤ n; 1 < j ≤ ci) holds ai, j > ai, j - 1.

In one swap operation you are allowed to choose two different cells of the table and swap the recorded there numbers, that is the number that was recorded in the first of the selected cells before the swap, is written in the second cell after it. Similarly,
the number that was recorded in the second of the selected cells, is written in the first cell after the swap.

Rearrange the numbers in the required manner. Note that you are allowed to perform any number of operations, but not more than s. You do not have to minimize
the number of operations.

Input

The first line contains a single integer n (1 ≤ n ≤ 50) that
shows the number of rows in the table. The second line contains n space-separated integers ci (1 ≤ ci ≤ 50; ci ≤ ci - 1) —
the numbers of cells on the corresponding rows.

Next n lines contain table а. The i-th
of them contains ci space-separated
integers: the j-th integer in this line represents ai, j.

It is guaranteed that all the given numbers ai, j are
positive and do not exceed s. It is guaranteed that all ai, j are
distinct.

Output

In the first line print a single integer m (0 ≤ m ≤ s),
representing the number of performed swaps.

In the next m lines print the description of these swap operations. In the i-th
line print four space-separated integers xi, yi, pi, qi (1 ≤ xi, pi ≤ n; 1 ≤ yi ≤ cxi; 1 ≤ qi ≤ cpi).
The printed numbers denote swapping the contents of cells axi, yi and api, qi.
Note that a swap operation can change the contents of distinct table cells. Print the swaps in the order, in which they should be executed.

Sample test(s)

input

3
3 2 1
4 3 5
6 1
2

output

2
1 1 2 2
2 1 3 1

input

1
4
4 3 2 1

output

2
1 1 1 4
1 2 1 3

注意到这么一句话：You do not have to minimize the number of operations.（你不必须使你的交换操作次数最小。）

所以对每个位置都从1~s编个号，只要数字不对应就交换到目标位置，详见代码。

完整代码：

/*30ms,132KB*/

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
#define X first
#define Y second
///上面的技巧很好~

vector<pair<pair<int, int>, pair<int, int> > > ans;
pair<int, int> num[2510];///这个数所在的坐标
int a[105][105], n, c[105];

int main(void)
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) 
        scanf("%d", &c[i]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= c[i]; j++)
		{
			scanf("%d", &a[i][j]);
			num[a[i][j]].X = i , num[a[i][j]].Y = j;
		}
    ////////////////////////
	int cnt = 0;
	///每个数只要不在位子上就交换到目标位置
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= c[i]; j++)
		{
			cnt++;
			if (a[i][j] == cnt)
				continue;
			ans.push_back(make_pair(make_pair(i, j), num[cnt]));
			int key = a[i][j];
			swap(a[i][j] , a[num[cnt].X][num[cnt].Y]);
			swap(num[key] , num[cnt]);
		}
	printf("%d\n", ans.size());
	for (int i = 0; i < ans.size(); i++)
		printf("%d %d %d %d\n" , ans[i].X.X , ans[i].X.Y , ans[i].Y.X , ans[i].Y.Y);
	return 0;
}