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 <iostream>#include <string>#include <vector>#include <algorithm>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" <<endl;                cout << "W" <<endl;                currentX--;            }        } 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" <<endl;                currentX++;            }        } else {            cerr << "now X same" <<endl;            if (currentY - lightY > 0){                cout << "N" << endl;                currentY--;            } else if (currentY - lightY < 0){                cout << "S" << endl;                currentY++;            }else {                // in position                cerr << "now Y same" <<endl;            }        }        // 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 <iostream>#include <string>#include <vector>#include <algorithm>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