題目描述

畫家小Q又開始他的藝術(shù)創(chuàng)作。小Q拿出了一塊有NxM像素格的畫板, 畫板初始狀態(tài)是空白的,用'X'表示。
小Q有他獨特的繪畫技巧,每次小Q會選擇一條斜線, 如果斜線的方向形如'/',即斜率為1,小Q會選擇這條斜線中的一段格子,都涂畫為藍色,用'B'表示;如果對角線的方向形如'',即斜率為-1,小Q會選擇這條斜線中的一段格子,都涂畫為黃色,用'Y'表示。
如果一個格子既被藍色涂畫過又被黃色涂畫過,那么這個格子就會變成綠色,用'G'表示。
小Q已經(jīng)有想畫出的作品的樣子, 請你幫他計算一下他最少需要多少次操作完成這幅畫。

輸入描述

每個輸入包含一個測試用例。
每個測試用例的第一行包含兩個正整數(shù)N和M(1 <= N, M <= 50), 表示畫板的長寬。
接下來的N行包含N個長度為M的字符串, 其中包含字符'B','Y','G','X',分別表示藍色,黃色,綠色,空白。整個表示小Q要完成的作品。

輸出描述

輸出一個正整數(shù), 表示小Q最少需要多少次操作完成繪畫。

示例1

輸入

4 4
YXXB
XYGX
XBYY
BXXY

輸出

3

說明

XXXX
XXXX
XXXX
XXXX
->
YXXX
XYXX
XXYX
XXXY
->
YXXB
XYBX
XBYX
BXXY
->
YXXB
XYGX
XBYY
BXXY

題目來源及自己的思路

鏈接

https://www.nowcoder.com/ques...

來源

?？途W(wǎng)

思路

從(0,0)位置開始遍歷

如果遍歷到(i,j)位置為B——>則繼續(xù)向左下方(i+1,j-1)和右上方(i-1,j+1)
遍歷時，遇到將B置為X，遇到G置為（消除B）置為Y
count++;
如果遍歷到(i,j)位置為Y——>則繼續(xù)向左上方(i-1,j-1)和右下方(i+1,j+1)
遍歷時，遇到將Y置為X，遇到G置為（消除Y）置為B
count++;
如果遍歷到(i,j)位置為G——>則分別執(zhí)行1,2

相關(guān)代碼

Java實現(xiàn)（能通過?。?/h2>

import java.util.Scanner; 
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String tem = scanner.nextLine();
        int n = Integer.valueOf(tem.split(" ")[0]);
        int m = Integer.valueOf(tem.split(" ")[1]);
        char color[][] = new char[n][m];
        for (int i = 0; i < n; i++) {
            tem = scanner.nextLine();
            for (int j = 0; j < tem.length(); j++) {
                color[i][j] = tem.charAt(j);
            }
        }
 
        getMinStep(n, m, color);
 
        scanner.close();
 
    }
 
    private static void getMinStep(int n, int m, char color[][]) {
 
        int step = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (color[i][j] == 'Y') {
                    dray(i, j, n, m, color); 
                    step++;
                } else if (color[i][j] == 'B') {
                    drab(i, j, n, m, color); 
                    step++;
                } else if (color[i][j] == 'G') {
                    dray(i, j, n, m, color); 
                    step++;
                    drab(i, j, n, m, color); 
                    step++;
                }
            }
        }
 
        System.out.println(step);
    }
    private static void dray(int x, int y, int n, int m, char color[][]) {
       
        if (x >= 0 && x < n && y >= 0 && y < m && (color[x][y] == 'Y' || color[x][y] == 'G')) {
            if (color[x][y] == 'G') {
                color[x][y] = 'B'; 
            } else {
                color[x][y] = 'X';
            }
 
            dray(x - 1, y - 1, n, m, color);
            dray(x + 1, y + 1, n, m, color);
 
        }
 
    }
 
    private static void drab(int x, int y, int n, int m, char color[][]) {
   
        if (x >= 0 && x < n && y >= 0 && y < m && (color[x][y] == 'B' || color[x][y] == 'G')) {
            if (color[x][y] == 'G') {
                color[x][y] = 'Y'; 
            } else {
                color[x][y] = 'X'; 
            }
 
            drab(x + 1, y - 1, n, m, color);
            drab(x - 1, y + 1, n, m, color);
 
        }
    }
 
}

JavaScript實現(xiàn)（不能通過?。〕瑫r?。?/h2>

while(line=readline()){
    let lines = line.split(" ");
    let n = parseInt(lines[0]);
    let m = parseInt(lines[1]);
    let arr = new Array();
    for(let i = 0; i < n; i++){
        let line2 = readline();
        arr[i] = new Array();
        for(let j = 0; j < m; j++){
            arr[i][j] = line2[j];
            //print(arr[i][j]);
        }
    }
    let step = 0;
    for(let i = 0; i < n; i++){
        for(let j = 0; j < m; j++){
            //print("---------## 訪問點(" + i +" , "+ j + " ): " +arr[i][j]);
            if(arr[i][j] == 'Y'){
                draw_y(arr, i, j, n, m);
                step++;
            }else if(arr[i][j] == 'B'){
                draw_b(arr, i, j, n, m);
                step++;
            }else if(arr[i][j] == 'G'){
                draw_y(arr, i, j, n, m);
                step++;
                draw_b(arr, i, j, n, m);
                step++
            }
        }
    }
    print(step);
}
function draw_y(arr,x,y,n,m){
    //print("draw_y:");
    //print(x,y,n,m);
    if(x >= 0 && x < n && y >=0 && y < m && (arr[x][y] == 'Y' || arr[x][y] == 'G')){
        if(arr[x][y] == 'Y'){
            arr[x][y] == 'X';
        }else{
            arr[x][y] == 'B';
        }
        //print("("+x+","+y+")"+"左上角：");
        draw_y(arr, x-1, y-1, n, m);
        
        //print("("+x+","+y+")"+"右下角：");
        draw_y(arr, x+1, y+1, n, m);
    }
}
function draw_b(arr,i,j,n,m){
    //print("draw_b:");
    //print(i,j,n,m);
    if(i >= 0 && i < n && j >=0 && j < m && (arr[i][j] == 'G' || arr[i][j] == 'B')){
        if(arr[i][j] == 'B'){
            arr[i][j] == 'X';
        }else{
            arr[i][j] == 'Y';
        }
        
        //print("("+i+","+j+")"+"左下角：");
        draw_b(arr, i-1, j+1, n, m);
        
        //print("("+i+","+j+")"+"右上角：");
        draw_b(arr, i+1,j-1,n,m);
    }
}

錯誤信息

已經(jīng)調(diào)試了好多天，就是無法通過！求助！

a.v8js:36: RangeError: Maximum call stack size exceeded
function draw_y(arr,x,y,n,m){
^
RangeError: Maximum call stack size exceeded
at draw_y (a.v8js:36:16)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)


你的輸出為:a.v8js:36: RangeError: Maximum call stack size exceeded
function draw_y(arr,x,y,n,m){
^
RangeError: Maximum call stack size exceeded
at draw_y (a.v8js:36:16)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)
at draw_y (a.v8js:49:9)
at draw_y (a.v8js:46:9)