Codeforces441C_Valera and Tubes(暴力)

Valera and Tubes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Valera has got a rectangle table consisting of n rows and m columns.
Valera numbered the table rows starting from one, from top to bottom and the columns – starting from one, from left to right. We will represent cell that is on the intersection of row x and
column y by a pair of integers (x, y).

Valera wants to place exactly k tubes on his rectangle table. A tube is such sequence of table cells (x1, y1), (x2, y2), ..., (xr, yr),
that:

• r ≥ 2;
• for any integer i (1 ≤ i ≤ r - 1) the following
  equation |xi - xi + 1| + |yi - yi + 1| = 1 holds;
• each table cell, which belongs to the tube, must occur exactly once in the sequence.

Valera thinks that the tubes are arranged in a fancy manner if the following conditions are fulfilled:

• no pair of tubes has common cells;
• each cell of the table belongs to some tube.

Help Valera to arrange k tubes on his rectangle table in a fancy manner.

Input

The first line contains three space-separated integers n, m, k (2 ≤ n, m ≤ 300; 2 ≤ 2k ≤ n·m)
— the number of rows, the number of columns and the number of tubes, correspondingly.

Output

Print k lines. In the i-th line print the description
of the i-th tube: first print integer ri (the
number of tube cells), then print 2ri integersxi1, yi1, xi2, yi2, ..., xiri, yiri (the
sequence of table cells).

If there are multiple solutions, you can print any of them. It is guaranteed that at least one solution exists.

Sample test(s)

input

3 3 3

output

3 1 1 1 2 1 3
3 2 1 2 2 2 3
3 3 1 3 2 3 3

input

2 3 1

output

6 1 1 1 2 1 3 2 3 2 2 2 1

Note

Picture for the first sample:

Picture for the second sample:

解题报告

让每一个人两个，剩下给第一个

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
int mmap[310][310];
int main()
{
    int n,m,k;
    while(cin>>n>>m>>k)
    {
        int t=n*m-(k-1)*2;
        int x=1,y=1;
        printf("%d",t);
        while(t--)
        {
            printf(" %d %d",x,y);
            mmap[x][y]=1;
            if(y+1<=m&&!mmap[x][y+1])
            {
                y++;
            }
            else if(y-1>=1&&!mmap[x][y-1])
                y--;
            else x++;
        }
        k--;
        while(k--)
        {
            printf("\n2");
            t=2;
            while(t--)
            {
                printf(" %d %d",x,y);
                mmap[x][y]=1;
                if(y+1<=m&&!mmap[x][y+1])
                {
                    y++;
                }
                else if(y-1>=1&&!mmap[x][y-1])
                    y--;
                else x++;
            }
        }
        printf("\n");
    }
}