codeforces Round #252 (Div. 2) C - Valera and Tubes

貪心演算法，每條路徑最短2格，故前k-1步每次走2格，最後一步全走完

由於資料比較小，可以先打表

#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
using namespace std;

typedef pair<int,int> Point;

int main(){
    int n, m, k, flag = -1;
    cin >> n >> m >>k;
    vector<Point> table;
    for(int i = 0 ; i < n ; ++i){
        flag*=-1;
        for(int j = 0; j < m; ++ j){
            table.push_back(Point(i,flag>0 ?  j : (m-1-j)));
        }
    }
    for(int i = 0 ; i <2*(k -1);  i+=2){
        cout<<2<<" "<<table[i].first+1<<" "<<table[i].second+1<<" "<<table[i+1].first+1<<" "<<table[i+1].second+1<<endl;
    }
    cout<<m*n-2*(k-1);
    for(int i = 2*(k-1); i < m*n; ++ i ){
        cout<<" "<<table[i].first+1<<" "<<table[i].second+1;
    }
    cout<<endl;
}

關於額外的資訊， |xi - xi + 1| + |yi - yi + 1| = 1，其實就是一個點的四連通區域，即上下左右4個點