題意。。。上ceoi官網看吧。。。

首先打一下sg函數發現必勝態和必敗態的分布位置是有規律的

于是我們只要知道最長步數的必勝態和最長步數的必敗態哪個更長就可以了

然后再打一下步數的表。。。發現必敗態的最長步數非常好確定，那么必勝態的步數搜一下就可以了！

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1 /************************************************************** 2 Problem: 1393 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:992 ms 7 Memory:2372 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 const int dx[] = {-2, -2, -1, 1}; 15 const int dy[] = {1, -1, -2, -2}; 16 const int N = 2e5 + 5; 17 const int inf = 1e8; 18 19 int n, k; 20 int Ans, ansx[N], ansy[N]; 21 22 inline int read(); 23 inline void print (const int &); 24 25 inline bool calc_sg(int x, int y) { 26 if ((x % 4 == 1 || x % 4 == 2) && (y % 4 == 1 || y % 4 == 2)) return 0; 27 if (n % 4 == 1 && ((x == n && y != n - 1) || (y == n && x != n - 1))) return 0; 28 if (n % 4 == 0 && x == n && y == n) return 0; 29 return 1; 30 } 31 32 inline bool in(const int &x, const int &y) { 33 return (x > 0 && y > 0 && x <= n && y <= n); 34 } 35 36 #define X x + dx[k] 37 #define Y y + dy[k] 38 inline int find_lose(int x, int y) { 39 if (x == n || y == n) return 2 * ((int) (x + y - 2) / 4); 40 return 2 * ((int) (x + y - 1) / 4); 41 } 42 43 inline int find_win(int x, int y) { 44 int k, res = 0; 45 for (k = 0; k < 4; ++k) 46 if (in(X, Y) && calc_sg(X, Y) == 0) 47 res = max(res, find_lose(X, Y)); 48 return res + 1; 49 } 50 51 inline int calc_lose(int x, int y, int i) { 52 int k, tmp = inf; 53 for (k = 0; k < 4; ++k) 54 if (in(X, Y) && calc_sg(X, Y) == 1 && find_win(X, Y) < tmp) { 55 tmp = find_win(X, Y); 56 ansx[i] = X, ansy[i] = Y; 57 } 58 } 59 60 inline int calc_win(int x, int y, int i) { 61 int k, tmp = -1; 62 for (k = 0; k < 4; ++k) 63 if (in(X, Y) && calc_sg(X, Y) == 0 && tmp < find_lose(X, Y)) { 64 tmp = find_lose(X, Y); 65 ansx[i] = X, ansy[i] = Y; 66 } 67 } 68 #undef X 69 #undef Y 70 71 int main() { 72 int i, x, y, tmp, max_lose = 0, max_win = 0; 73 k = read(), n = read(); 74 for (i = 1; i <= k; ++i) { 75 x = read(), y = read(); 76 tmp = calc_sg(x, y); 77 if (tmp == 0) { 78 max_lose = max(max_lose, find_lose(x, y)); 79 calc_lose(x, y, i); 80 continue; 81 } 82 max_win = max(max_win, find_win(x, y)); 83 calc_win(x, y, i); 84 } 85 if (max_lose > max_win) puts("NO"); else { 86 puts("YES"); 87 for (i = 1; i <= k; ++i) { 88 print(ansx[i]), putchar(' '); 89 print(ansy[i]), putchar('\n'); 90 } 91 } 92 return 0; 93 } 94 95 inline int read() { 96 int x = 0; 97 char ch = getchar(); 98 while (ch < '0' || '9' < ch) 99 ch = getchar(); 100 while ('0' <= ch && ch <= '9') { 101 x = x * 10 + ch - '0'; 102 ch = getchar(); 103 } 104 return x; 105 } 106 107 inline void print(const int &x) { 108 static int tot, pr[20], t; 109 t = x, tot = 0; 110 while (t) 111 pr[++tot] = t % 10, t /= 10; 112 if (!tot) putchar('0'); 113 while(tot) 114 putchar('0' + pr[tot--]); 115 } View Code

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轉載于:https://www.cnblogs.com/rausen/p/4352029.html

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