在第一章我们已经说了怎么才能“夹一个”以及怎样才能挑一对,但那毕竟只是书面上的,对码农来讲,我们还是用代码讲解起来会更容易了解。
为了更容易对照分析,我们先把路线再次贴出来:
// 可走的路线
this.lines = [
[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[ 0, 5, 10, 15, 20],
[ 1, 6, 11, 16, 21],
[ 2, 7, 12, 17, 22],
[ 3, 8, 13, 18, 23],
[ 4, 9, 14, 19, 24],
[ 0, 6, 12, 18, 24],
[ 4, 8, 12, 16, 20],
[ 2, 6, 10],
[ 2, 8, 14],
[10, 16, 22],
[14, 18, 22]
];
一、夹一个:
根据上面给出的有限路线中,要实现“夹一个”,首页这颗棋子的index得在[0,1,2...24]之中,我们循环搜索每条路线,只要找出符合条件的路线和位置就可把对方的棋子给吃掉。
首先我们找出棋子的目标位置是在哪条路线中:
int index = $.inArray(chess.point.index, this.lines[i]);//chess被移动的棋子,下同
if(index!=-1)//...
然后再找出该条线上能被吃掉的棋子是哪一个。如果按照水平方向来看,被吃掉的棋子有可能在左边,也有可能在右边,如果在左边,那么该方还有一个棋子应该在被吃掉的棋子的左边:
var p1 = index > 1 ? this.chesses[this.lines[i][index - 1]].player : Player.None;
if (p1 != Player.None && p1 != chess.player) {
if (this.chesses[this.lines[i][index - 2]].player == chess.player) {
//...找到符合条件的路线
同理,如果被吃掉的棋子在右边,那么该方还有一个棋子应该在被吃掉的棋子的右边:
var p2 = index < this.lines[i].length - 2 ? this.chesses[this.lines[i][index + 1]].player : Player.None;
if (p2 != Player.None && p2 != chess.player) {
if (this.chesses[this.lines[i][index + 2]].player == chess.player) {
//...找到符合条件的路线
不过,因为按照规则,能夹对方棋子的同时,该条路径上仅且只能有三颗棋子,已方两颗,对方一颗,其他位置上是不能有棋子存在的:
对于在左边的情况:
var bfind = true;// 是否找到能被吃的棋子
for (j = 0; j < index - 2; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) return;
bfind = true;
for (j = index + 1; j < this.lines[i].length; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) return;
chessArray.push([this.chesses[this.lines[i][index - 1]]]);// 找到了
对于在右边的情况:
var bfind = true;
for (j = 0; j < index; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) return;
bfind = true;
for (j = index + 3; j < this.lines[i].length; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) return;
chessArray.push([this.chesses[this.lines[i][index + 1]]]);// 找到了
对于找到可以被夹掉的棋子我们记录下来,存到 chessArray 里面,以便进行其他操作。
二、挑一对:
同样,我们先找出棋子在哪条路径中:
index = $.inArray(chess.point.index, this.lines[i]);
if (index > 0 && index < this.lines[i].length - 1) {
然后相对于夹一个来说简单很多,我们只要找出该棋子左右相邻的两个棋子是对方的棋子,且该条直线上其他位置都是空位就行了。
先找出左右相邻的两颗棋子:
var p1 = this.chesses[this.lines[i][index - 1]].player;
var p2 = this.chesses[this.lines[i][index + 1]].player;
if (p1 != chess.player && p2 != chess.player && p1 != Player.None && p2 != Player.None) {
再判断其他位置是空位:
var bfind = true;
for (j = 0; j < index - 1; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) return;
bfind = true;
for (j = this.lines[i].length - 1; j > index + 1; j--) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) return;
chessArray.push([this.chesses[this.lines[i][index - 1]], this.chesses[this.lines[i][index + 1]]]);// 找到了
现在实现了两个基本的函数,下一章里沃特再跟大家分析移动棋子。本章实现的两个函数归纳如下:
// 是否可“挑一对”
this.canCarry = function (chess) {
var p1, p2, j, index, bfind, chessArray = [];
for (var i = 0; i < this.lines.length; i++) {
index = $.inArray(chess.point.index, this.lines[i]);
if (index > 0 && index < this.lines[i].length - 1) {
p1 = this.chesses[this.lines[i][index - 1]].player;
p2 = this.chesses[this.lines[i][index + 1]].player;
if (p1 != chess.player && p2 != chess.player && p1 != Player.None && p2 != Player.None) {
bfind = true;
for (j = 0; j < index - 1; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) continue;
bfind = true;
for (j = this.lines[i].length - 1; j > index + 1; j--) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) continue;
chessArray.push([this.chesses[this.lines[i][index - 1]], this.chesses[this.lines[i][index + 1]]]);
}
}
}
return chessArray.length == 0 ? false : chessArray;
};
// 是否可“夹一个”
this.canClip = function (chess) {
var p1, p2, j, index, bfind, chessArray = [];
for (var i = 0; i < this.lines.length; i++) {
index = $.inArray(chess.point.index, this.lines[i]);
if (index != -1) {
p1 = index > 1 ? this.chesses[this.lines[i][index - 1]].player : Player.None;
p2 = index < this.lines[i].length - 2 ? this.chesses[this.lines[i][index + 1]].player : Player.None;
if (p1 != Player.None && p1 != chess.player) {
if (this.chesses[this.lines[i][index - 2]].player == chess.player) {
bfind = true;
for (j = 0; j < index - 2; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) continue;
bfind = true;
for (j = index + 1; j < this.lines[i].length; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) continue;
chessArray.push([this.chesses[this.lines[i][index - 1]]]);
}
} else if (p2 != Player.None && p2 != chess.player) {
if (this.chesses[this.lines[i][index + 2]].player == chess.player) {
bfind = true;
for (j = 0; j < index; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) continue;
bfind = true;
for (j = index + 3; j < this.lines[i].length; j++) {
if (this.chesses[this.lines[i][j]].player != Player.None) { bfind = false; break; }
}
if (!bfind) continue;
chessArray.push([this.chesses[this.lines[i][index + 1]]]);
}
}
}
}
return chessArray.length == 0 ? false : chessArray;
};