标签：red 约束 ace log under pad height value none
| 描述 | |
| 在一个N×N的棋盘上下五子棋，给定一个五子棋黑白棋的落子序列 (x0,y0),(x1,y1), ..., （xn,yn），判断走到多少步时，哪方获胜。 | |
| 关于输入 | |
| 第一行有两个整数，棋盘的大小N和落子序列的长度n。 其余各行每行两个数字，分别表示黑棋或白棋棋子在棋盘上的位置。 定输入一定合理的，而且一定能分出输赢，但因为对弈双方没有判断出输赢而多行了数步。 | |
| 关于输出 | |
| 输出为一行，包含两个整数，第一个整数是判断出输赢时行棋的步数，第二个数字表示获胜方。如果黑方获胜，输出0；如果白方获胜，输出1。 | |
| 例子输入 | |
10 12
0 0
1 0
0 1
2 0
0 2
3 0
0 3
4 0
0 4
5 0
0 5
6 0
|
|
| 例子输出 | |
9 0
|
|
| 提示 | |
| 五子棋规则比较复杂，这里我们只要求黑棋先行，谁先连出至少5连子就算赢，不管是否出现规则中约束的禁手。每局棋都有胜负结果。 |
1 #include<iostream> 2 using namespace std; 3 int main() 4 { 5 int N, n,flat=0; 6 cin >> N >> n; 7 int chess[50][50]; 8 int col1, line1,col2,line2; 9 for (int i = 0; i < 50; i++) 10 for (int j = 0; j < 50; j++) 11 chess[i][j] = 0; 12 for (int i = 1; i <= n; i++) 13 { 14 if (i % 2 != 0) 15 { 16 cin >> line1 >> col1; 17 chess[line1][col1] = -1; 18 for (int j = 0; j <= 4; j++) 19 { 20 if (line1 + j > N || col1 + j > N) 21 continue; 22 if (line1 - 4 + j < 0 || col1 - 4 + j < 0) 23 continue; 24 flat = 0; 25 for (int p = 0; p <= j; p++) 26 { 27 if (chess[line1 + p][col1 + p] == -1) 28 flat++; 29 } 30 for (int p = 0; p <= 4 - j; p++) 31 { 32 if (chess[line1 - p][col1 - p] == -1) 33 flat++; 34 } 35 if (flat == 6) 36 { 37 cout << i << " " << "0"; 38 return 0; 39 } 40 } 41 flat = 0; 42 for (int j = 0; j <= 4; j++) 43 { 44 if (line1 + j > N || col1 - j < 0) 45 continue; 46 if (line1 - 4 + j < 0 || col1 + 4 - j >N) 47 continue; 48 flat = 0; 49 for (int p = 0; p <= j; p++) 50 { 51 if (chess[line1 + p][col1 - p] == -1) 52 flat++; 53 } 54 for (int p = 0; p <= 4 - j; p++) 55 { 56 if (chess[line1 - p][col1 + p] == -1) 57 flat++; 58 } 59 if (flat == 6) 60 { 61 cout << i << " " << "0"; 62 return 0; 63 } 64 } 65 flat = 0; 66 for (int j = 0; j <= 4; j++) 67 { 68 if (line1 + j > N) 69 continue; 70 if (line1 - 4 + j < 0) 71 continue; 72 flat = 0; 73 for (int p = 0; p <= j; p++) 74 { 75 if (chess[line1 + p][col1] == -1) 76 flat++; 77 } 78 for (int p = 0; p <= 4 - j; p++) 79 { 80 if (chess[line1 - p][col1] == -1) 81 flat++; 82 } 83 if (flat == 6) 84 { 85 cout << i << " " << "0"; 86 return 0; 87 } 88 } 89 flat = 0; 90 for (int j = 0; j <= 4; j++) 91 { 92 if (col1 + j > N) 93 continue; 94 if (col1 - 4 + j < 0) 95 continue; 96 flat = 0; 97 for (int p = 0; p <= j; p++) 98 { 99 if (chess[line1][col1 + p] == -1) 100 flat++; 101 } 102 for (int p = 0; p <= 4 - j; p++) 103 { 104 if (chess[line1][col1 - p] == -1) 105 flat++; 106 } 107 if (flat == 6) 108 { 109 cout << i << " " << "0"; 110 return 0; 111 } 112 } 113 } 114 else 115 { 116 cin >> line2 >> col2; 117 chess[line2][col2] =1; 118 flat = 0; 119 for (int j = 0; j <= 4; j++) 120 { 121 if (line2 + j > N || col2 + j > N) 122 continue; 123 if (line2 - 4 + j < 0 || col2 - 4 + j < 0) 124 continue; 125 flat = 0; 126 for (int p = 0; p <= j; p++) 127 { 128 if (chess[line2 + p][col2 + p] == 1) 129 flat++; 130 } 131 for (int p = 0; p <= 4 - j; p++) 132 { 133 if (chess[line2 - p][col2 - p] == 1) 134 flat++; 135 } 136 if (flat ==6) 137 { 138 cout << i << " " << "0"; 139 return 0; 140 } 141 } 142 flat = 0; 143 for (int j = 0; j <= 4; j++) 144 { 145 if (line2 + j > N || col2 - j < 0) 146 continue; 147 if (line2 - 4 + j < 0 || col2 + 4 - j >N) 148 continue; 149 flat = 0; 150 for (int p = 0; p <= j; p++) 151 { 152 if (chess[line2 + p][col2 - p] == 1) 153 flat++; 154 } 155 for (int p = 0; p <= 4 - j; p++) 156 { 157 if (chess[line2 - p][col2 + p] == 1) 158 flat++; 159 } 160 if (flat == 6) 161 { 162 cout << i << " " << "0"; 163 return 0; 164 } 165 } 166 flat = 0; 167 for (int j = 0; j <= 4; j++) 168 { 169 if (line2 + j > N) 170 continue; 171 if (line2 - 4 + j < 0) 172 continue; 173 flat = 0; 174 for (int p = 0; p <= j; p++) 175 { 176 if (chess[line2 + p][col2] == 1) 177 flat++; 178 } 179 for (int p = 0; p <= 4 - j; p++) 180 { 181 if (chess[line2 - p][col2] == 1) 182 flat++; 183 } 184 if (flat == 6) 185 { 186 cout << i << " " << "0"; 187 return 0; 188 } 189 } 190 flat = 0; 191 for (int j = 0; j <= 4; j++) 192 { 193 if (col2 + j > N) 194 continue; 195 if (col2 - 4 + j < 0) 196 continue; 197 flat = 0; 198 for (int p = 0; p <= j; p++) 199 { 200 if (chess[line2][col2 + p] == 1) 201 flat++; 202 } 203 for (int p = 0; p <= 4 - j; p++) 204 { 205 if (chess[line2][col2 - p] == 1) 206 flat++; 207 } 208 if (flat ==6) 209 { 210 cout << i << " " << "0"; 211 return 0; 212 } 213 } 214 } 215 } 216 return 0; 217 }
在我看来这道题非常考验耐心和细心……以及如果是考试的话一定要冷静……
不知道有没有更便捷些的解法
（神烦
标签：red 约束 ace log under pad height value none
原文地址：http://www.cnblogs.com/yalphait/p/7779222.html
