【Codeforces 237B Young Table】+ map 数组构造

B. Young Table
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) holds ci ≤ 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 the i-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.
Examples
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

把原来给出的数组按给出的要求排序：
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.
为需要交换的次数以及需要交换的位置坐标~~
把数组里的元素升序排列~枚举原数组里的元素~不一样的换~并记录位置~位置记录可用 map + pair 实现比较方便~~

AC代码：

#include<cstdio>#include<map>#include<algorithm>using namespace std;typedef pair <int,int> p;int ma[60],mp[60][60],n[4010][4],pa[4010],N;map <int,p> m;int main(){    int M = 0;    scanf("%d",&N);    for(int i = 1 ; i <= N; i++)        scanf("%d",&ma[i]);    for(int i = 1 ; i <= N; i++)       for(int j = 1 ; j <= ma[i] ; j++){           scanf("%d",&mp[i][j]);           m[mp[i][j]] = p(i,j); // 建立映射关系           pa[++M] = mp[i][j];    }    sort(pa + 1, pa + 1 + M);    int kl = 0, pl = 0;    for(int i = 1 ; i <= N; i++)        for(int j = 1 ; j <= ma[i] ; j++){            if(mp[i][j] != pa[++pl]){ // 需要换位置，记录位置坐标                n[++kl][0] = i;                n[kl][1] = j;                int x = m[pa[pl]].first;                int y = m[pa[pl]].second;                n[kl][2] = x;                n[kl][3] = y;                m[mp[i][j]] = p (x,y); // 之前的坐标用不到了，但换过去的坐标要记得换                swap(mp[i][j],mp[x][y]);            }    }    printf("%d\n",kl);    for(int i = 1 ; i <= kl ; i++)        printf("%d %d %d %d\n",n[i][0],n[i][1],n[i][2],n[i][3]);    return 0;}