利用缓冲区实现“向量分解” - Power of Thor - Episode 1 [CodingGame技巧总结]

问题描述

Your program must allow Thor to reach the light of power.

题目地址：
https://www.codingame.com/ide/puzzle/power-of-thor-episode-1

原始解法

通过嵌套的if条件语句来枚举各类情况：

#include 
#include 
#include 
#include 

using namespace std;

/**
 * Auto-generated code below aims at helping you parse
 * the standard input according to the problem statement.
 * ---
 * Hint: You can use the debug stream to print initialTX and initialTY, if Thor seems not follow your orders.
 **/
int main()
{
    int lightX; // the X position of the light of power
    int lightY; // the Y position of the light of power
    int initialTX; // Thor's starting X position
    int initialTY; // Thor's starting Y position
    cin >> lightX >> lightY >> initialTX >> initialTY; cin.ignore();

    int currentX = initialTX;
    int currentY = initialTY;

    // game loop
    while (1) {
        int remainingTurns; // The remaining amount of turns Thor can move. Do not remove this line.
        cin >> remainingTurns; cin.ignore();

        // Write an action using cout. DON'T FORGET THE "<< endl"
        // To debug: cerr << "Debug messages..." << endl;


        // if both offset, then move in diaganol line
        if (currentX - lightX > 0) {
            if (currentY - lightY > 0){
                cout << "NW" << endl;
                currentX--;
                currentY--;
            } else if (currentY - lightY <0){
                cout << "SW" << endl;

                currentX--;
                currentY++;
                cerr << currentX << endl;
            }else {
                cerr << "now Y same" <cout << "W" <else if (currentX - lightX < 0) {
            if (currentY - lightY > 0){
                cout << "NE" << endl;
                currentX++;
                currentY--;
            } else if (currentY - lightY <0){
                cout << "SE" << endl;
                currentX++;
                currentY++;
            }else {
                cout << "E" <else {
            cerr << "now X same" <if (currentY - lightY > 0){
                cout << "N" << endl;
                currentY--;
            } else if (currentY - lightY < 0){
                cout << "S" << endl;
                currentY++;
            }else {
                // in position
                cerr << "now Y same" <// A single line providing the move to be made: N NE E SE S SW W or NW
        // cout << "SE" << endl;
    }
}

“向量分解”解法

#include 
#include 
#include 
#include 

using namespace std;

/**
 * Auto-generated code below aims at helping you parse
 * the standard input according to the problem statement.
 * ---
 * Hint: You can use the debug stream to print initialTX and initialTY, if Thor seems not follow your orders.
 **/
int main()
{
    int lightX; // the X position of the light of power
    int lightY; // the Y position of the light of power
    int initialTX; // Thor's starting X position
    int initialTY; // Thor's starting Y position
    cin >> lightX >> lightY >> initialTX >> initialTY; cin.ignore();

    int dx = lightX - initialTX;
    int dy = lightY - initialTY;

    // game loop
    while (1) {
        int remainingTurns;
        cin >> remainingTurns; cin.ignore();

        // Write an action using cout. DON'T FORGET THE "<< endl"
        // To debug: cerr << "Debug messages..." << endl;
        if (dy > 0) {
            cout << "S";
            dy--;
        }           
        if (dy < 0) {
            cout << "N";
            dy++;
        }  

        if (dx > 0) {
            cout << "E";
            dx--;
        }           
        if (dx < 0) {
            cout << "W";
            dx++;
        }          

        cout << endl; // A single line providing the move to be made: N NE E SE S SW W or NW
    }
}

技巧分析

在物理学中可以对速度向量进行向量分解，最常用的是分解为X、Y方向两个速度分量。

在C++实现中，虽然我们不能简单引入速度这个概念，但是通过缓冲区的暂存，可以让我们通过单独判定X、Y方向的关系，分别向缓冲区中写入内容，最后用endl一并输出。

潜在问题是如果向缓冲区中写出过多字符，可能因为缓冲区已满而自动输出了。应该可以通过临时变量来解决。

技巧泛化

当输出结果与判定条件都可以进行某种对应形式的分解时，可以适当利用缓冲区（输出缓冲区，或者临时变量）达到分解的目的，从而简化程序逻辑并提高效率。

参考

CodingGame ，ower of Thor - Episode 1，C++最高票答案，2017.02.07