求黑白棋算法。 求C# 黑白棋翻转算法
\u6c42java\u9ed1\u767d\u68cb\u641c\u7d22\u7b97\u6cd5\u3002\u3002\u3002\u3002public boolean fanqi(int x, int y) {
boolean a = false;
if (qi[x][y] == 0) {
if (flag) {// \u9ed1\u5403\u767d
if (y - 1 > 0 && qi[x][y - 1] == -1) {
for (int i = 2; y - i >= 0; i++) {
if (qi[x][y - i] == 1) {
for (int n = 1; n < i; n++) {
qi[x][y - n] = 1;
}
a = true;
if (qi[x][y - i] == 0)
i = 9;
}
}
}
if (y + 1 < 8 && qi[x][y + 1] == -1) {
for (int i = 2; y + i < 8; i++) {
if (qi[x][y + i] == 1) {
for (int n = 1; n < i; n++) {
qi[x][y + n] = 1;
}
a = true;
if (qi[x][y + i] == 0)
i = 9;
}
}
}
if (x + 1 < 8 && qi[x + 1][y] == -1) {
for (int i = 2; x + i < 8; i++) {
if (qi[x + i][y] == 1) {
for (int n = 1; n < i; n++) {
qi[x + n][y] = 1;
}
a = true;
if (qi[x + i][y] == 0)
i = 9;
}
}
}
if (x - 1 >= 0 && qi[x - 1][y] == -1) {
for (int i = 2; x - i >= 0; i++) {
if (qi[x - i][y] == 1) {
for (int n = 1; n < i; n++) {
qi[x - n][y] = 1;
}
a = true;
if (qi[x - i][y] == 0)
i = 9;
}
}
}
if (x - 1 >= 0 && y - 1 >= 0 && qi[x - 1][y - 1] == -1) {
for (int i = 2; x - i >= 0 && y - i >= 0; i++) {
if (qi[x - i][y - i] == 1) {
for (int n = 1; n < i; n++) {
qi[x - n][y - n] = 1;
}
a = true;
if (qi[x - i][y - i] == 0)
i = 9;
}
}
}
if (x + 1 < 8 && y - 1 >= 0 && qi[x + 1][y - 1] == -1) {
for (int i = 2; x + i < 8 && y - i >= 0; i++) {
if (qi[x + i][y - i] == 1) {
for (int n = 1; n < i; n++) {
qi[x + n][y - n] = 1;
}
a = true;
if (qi[x + i][y - i] == 0)
i = 9;
}
}
}
if (x - 1 >= 0 && y + 1 < 8 && qi[x - 1][y + 1] == -1) {
for (int i = 2; x - i >= 0 && y + i < 8; i++) {
if (qi[x - i][y + i] == 1) {
for (int n = 1; n < i; n++) {
qi[x - n][y + n] = 1;
}
a = true;
if (qi[x - i][y + i] == 0)
i = 9;
}
}
}
if (x + 1 < 8 && y + 1 < 8 && qi[x + 1][y + 1] == -1) {
for (int i = 2; x + i < 8 && y + i < 8; i++) {
if (qi[x + i][y + i] == 1) {
for (int n = 1; n < i; n++) {
qi[x + n][y + n] = 1;
}
a = true;
if (qi[x + i][y + i] == 0)
i = 9;
}
}
}
}
else {// \u767d\u5403\u9ed1
if (y - 1 > 0 && qi[x][y - 1] == 1) {
for (int i = 2; y - i >= 0; i++) {
if (qi[x][y - i] == -1) {
for (int n = 1; n < i; n++) {
qi[x][y - n] = -1;
}
a = true;
if (qi[x][y - i] == 0)
i = 9;
}
}
}
if (y + 1 < 8 && qi[x][y + 1] == 1) {
for (int i = 2; y + i < 8; i++) {
if (qi[x][y + i] == -1) {
for (int n = 1; n < i; n++) {
qi[x][y + n] = -1;
}
a = true;
if (qi[x][y + i] == 0)
i = 9;
}
}
}
if (x + 1 < 8 && qi[x + 1][y] == 1) {
for (int i = 2; x + i < 8; i++) {
if (qi[x + i][y] == -1) {
for (int n = 1; n < i; n++) {
qi[x + n][y] = -1;
}
a = true;
if (qi[x + i][y] == 0)
i = 9;
}
}
}
if (x - 1 >= 0 && qi[x - 1][y] == 1) {
for (int i = 2; x - i >= 0; i++) {
if (qi[x - i][y] == -1) {
for (int n = 1; n < i; n++) {
qi[x - n][y] = -1;
}
a = true;
if (qi[x - i][y] == 0)
i = 9;
}
}
}
if (x - 1 >= 0 && y - 1 >= 0 && qi[x - 1][y - 1] == 1) {
for (int i = 2; x - i >= 0 && y - i >= 0; i++) {
if (qi[x - i][y - i] == -1) {
for (int n = 1; n < i; n++) {
qi[x - n][y - n] = -1;
}
a = true;
if (qi[x - i][y - i] == 0)
i = 9;
}
}
}
if (x + 1 < 8 && y - 1 >= 0 && qi[x + 1][y - 1] == 1) {
for (int i = 2; x + i < 8 && y - i >= 0; i++) {
if (qi[x + i][y - i] == -1) {
for (int n = 1; n < i; n++) {
qi[x + n][y - n] = -1;
}
a = true;
if (qi[x + i][y - i] == 0)
i = 9;
}
}
}
if (x - 1 >= 0 && y + 1 < 8 && qi[x - 1][y + 1] == 1) {
for (int i = 2; x - i >= 0 && y + i < 8; i++) {
if (qi[x - i][y + i] == -1) {
for (int n = 1; n < i; n++) {
qi[x - n][y + n] = -1;
}
a = true;
if (qi[x - i][y + i] == 0)
i = 9;
}
}
}
if (x + 1 < 8 && y + 1 < 8 && qi[x + 1][y + 1] == 1) {
for (int i = 2; x + i < 8 && y + i < 8; i++) {
if (qi[x + i][y + i] == -1) {
for (int n = 1; n < i; n++) {
qi[x + n][y + n] = -1;
}
a = true;
if (qi[x + i][y + i] == 0)
i = 9;
}
}
}
}
}
if (a)
return true;
else
return false;
}
\u5982\u679c\u662f\u4fee\u6539\u73b0\u6709\u4ee3\u7801\uff0c\u53ef\u4ee5\u7559\u8054\u7cfb\u77a7\u77a7\u3002
优先级算法(1-8)x( a-h)对这64个格分别定义优先级，优先级高的先落子，同优先级的（同优先级的格子非常少）考虑吃子数。
算法简单极易实现，做的AI保证无敌。
当然 这64个格子的具体优先级，需要精通黑白棋的人来分析了。
斑马的算法是公开的，去看斑马的源代码去 。顺便吐槽一下那个提出优先级算法的。黑白棋中的星位是拥有大攻击力和大损害的点位，只有在合适的时机去踩才是好点，按照优先级来，基本就可以跳过这个点位了，那么你的棋力也有限得紧了。其实现在就是拼计算机的质量，老老实实穷举是最无敌的，以前的破机器都能反搜16步，拼着巨型机的性能，反搜30步也不是不行，做好剪枝和平衡的话，我觉得电脑已经无敌了
绛旓細浼樺厛绾х畻娉(1-8)x( a-h)瀵硅繖64涓牸鍒嗗埆瀹氫箟浼樺厛绾锛屼紭鍏堢骇楂樼殑鍏堣惤瀛愶紝鍚屼紭鍏堢骇鐨勶紙鍚屼紭鍏堢骇鐨勬牸瀛愰潪甯稿皯锛夎冭檻鍚冨瓙鏁般傜畻娉曠畝鍗曟瀬鏄撳疄鐜帮紝鍋氱殑AI淇濊瘉鏃犳晫銆傚綋鐒 杩64涓牸瀛愮殑鍏蜂綋浼樺厛绾э紝闇瑕佺簿閫氶粦鐧芥鐨勪汉鏉ュ垎鏋愪簡銆
绛旓細4x+18=2*锛3x+1锛夎В寰梮=8 鍗8娆″悗锛岀櫧浣1锛岄粦浣18
绛旓細public boolean fanqi(int x, int y) { boolean a = false;if (qi[x][y] == 0) { if (flag) {// 榛戝悆鐧 if (y - 1 > 0 && qi[x][y - 1] == -1) { for (int i = 2; y - i >= 0; i++) { if (qi[x][y - i] == 1) { for (int n = 1; n < ...
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