[USACO 6.4.1] The Primes
题目大意
给一个5*5的矩阵,在每个格子里面填入0~9这10个数字.要求行从左往右读,列从上往下读,两条对角线从上往下读是一个5位素数且不含前导0.
现在给出了开头的数字和行,列,对角线之和,求符合规定的所有数字矩阵.
题解
这是一道搜索题.
但盲目的搜索是会TLE的.因此,我们应当采用合理的剪枝.
由题目可知,加入行列对角线我们知道了任意4位数,我们可以推出第5位数是什么.
因此,我们可以采用传说中的手动指定搜索顺序的方法来搜索和回溯.
具体搜索顺序如下图.
大概就是这样的感觉.
里面的C是可以直接算出来的.所以我们只需要搜索只有数字那些就行了.
还有我一开始是枚举对角线素数.
注意: 先枚举每个数的最后一位会让你减少至少一半的搜索量.因为5位素数的最后一位数肯定不是偶数.
详情请看代码.代码有点恶心.我用了宏定义,因为这样看起来会更加恶心.
代码
/* TASK:prime3 LANG:C++ */ #include <cstdio> #include <cstring> #include <algorithm> #define REP(i,st,ed) for ((i) = (st); (i) < (ed); ++(i)) using namespace std; struct Ans { int matrix[5][5]; Ans operator = (Ans &rhs) { memcpy(matrix, rhs.matrix, sizeof(rhs.matrix)); return *this; } bool operator < (const Ans &rhs) const { for (int i = 0; i < 5; ++i) for (int j = 0; j < 5; ++j) if (matrix[i][j] < rhs.matrix[i][j]) return true; else if (matrix[i][j] > rhs.matrix[i][j]) return false; return false; } }ans[1000], tmp; int anslen; const int MAXP = 100000; int p[10000], np, sumallr[5], sumallc[5], sumalld, sum; bool v[MAXP * 10]; bool check(int x) { int s = 0; while (x) s += x % 10, x /= 10; return s == sum; } void filtrate() { memset(v, true, sizeof(v)); for (int i = 2; i < MAXP; ++i) for (int j = 2; i * j < MAXP; ++j) v[i * j] = false; np = 0; for (int i = 10000; i < MAXP; ++i) if (v[i] && check(i)) p[np++] = i; } bool judge(int x, int y) { if (sumallr[x] + tmp.matrix[x][y] > sum) return false; if (sumallc[y] + tmp.matrix[x][y] > sum) return false; if (4 - x == y && sumalld + tmp.matrix[x][y] > sum) return false; return true; } void adds(int x, int y, int d) { sumallr[x] += tmp.matrix[x][y] * d; sumallc[y] += tmp.matrix[x][y] * d; if (4 - x == y) sumalld += tmp.matrix[x][y] * d; } int trans(int d, int m) { if (m == 1) return tmp.matrix[d][0] * 10000 + tmp.matrix[d][1] * 1000 + tmp.matrix[d][2] * 100 + tmp.matrix[d][3] * 10 + tmp.matrix[d][4]; else if (m == 2) return tmp.matrix[0][d] * 10000 + tmp.matrix[1][d] * 1000 + tmp.matrix[2][d] * 100 + tmp.matrix[3][d] * 10 + tmp.matrix[4][d]; else if (m == 3) return tmp.matrix[4][0] * 10000 + tmp.matrix[3][1] * 1000 + tmp.matrix[2][2] * 100 + tmp.matrix[1][3] * 10 + tmp.matrix[0][4]; } bool condit(int x) { return x >= 0 && x <= 9; } int main() { freopen("prime3.in", "r", stdin); freopen("prime3.out", "w", stdout); scanf("%d%d", &sum, &tmp.matrix[0][0]); filtrate(); memset(sumallr, 0, sizeof(sumallr)); memset(sumallc, 0, sizeof(sumallc)); sumalld = 0; anslen = 0; for (int i = 0; i < np; ++i) { if (p[i] * 10 / MAXP != tmp.matrix[0][0]) continue; int k = p[i]; for (int j = 4; j >= 0; --j) { tmp.matrix[j][j] = k % 10; sumallr[j] = sumallc[j] = k % 10; k /= 10; } sumalld = p[i] % 1000 / 100; REP(tmp.matrix[0][4], 1, 10) { if (tmp.matrix[0][4] % 2 == 0 || tmp.matrix[0][4] == 5) continue; if (!judge(0, 4)) break; adds(0, 4, 1); REP(tmp.matrix[1][3], 0, 10) { if (!judge(1, 3)) break; adds(1, 3, 1); REP(tmp.matrix[4][0], 1, 10) { if (tmp.matrix[4][0] % 2 == 0 || tmp.matrix[4][0] == 5) continue; if (!judge(4, 0)) break; tmp.matrix[3][1] = sum - sumalld - tmp.matrix[4][0]; if (!condit(tmp.matrix[3][1]) || !judge(3, 1)) continue; if (!v[trans(0, 3)]) continue; adds(3, 1, 1); adds(4, 0, 1); REP(tmp.matrix[1][4], 1, 10) { if (tmp.matrix[1][4] % 2 == 0 || tmp.matrix[1][4] == 5) continue; if (!judge(1, 4)) break; adds(1, 4, 1); REP(tmp.matrix[2][4], 1, 10) { if (tmp.matrix[2][4] % 2 == 0 || tmp.matrix[2][4] == 5) continue; if (!judge(2, 4)) break; tmp.matrix[3][4] = sum - sumallc[4] - tmp.matrix[2][4]; if (!condit(tmp.matrix[3][4]) || !judge(3, 4)) continue; if (!v[trans(4, 2)]) continue; adds(3, 4, 1); adds(2, 4, 1); REP(tmp.matrix[4][3], 1, 10) { if (tmp.matrix[4][3] % 2 == 0 || tmp.matrix[4][3] == 5) continue; if (!judge(4, 3)) break; adds(4, 3, 1); REP(tmp.matrix[0][3], 1, 10) { if (!judge(0, 3)) break; tmp.matrix[2][3] = sum - sumallc[3] - tmp.matrix[0][3]; if (!condit(tmp.matrix[2][3]) || !judge(2, 3)) continue; if (!v[trans(3, 2)]) continue; adds(0, 3, 1); adds(2, 3, 1); REP(tmp.matrix[4][2], 1, 10) { if (tmp.matrix[4][2] % 2 == 0 || tmp.matrix[4][2] == 5) continue; if (!judge(4, 2)) break; adds(4, 2, 1); REP(tmp.matrix[3][2], 0, 10) { if (!judge(3, 2)) break; adds(3, 2, 1); REP(tmp.matrix[0][2], 1, 10) { if (!judge(0, 2)) break; tmp.matrix[1][2] = sum - sumallc[2] - tmp.matrix[0][2]; if (!condit(tmp.matrix[1][2]) || !judge(1, 2) || !v[trans(2, 2)]) continue; tmp.matrix[0][1] = sum - sumallr[0] - tmp.matrix[0][2]; if (tmp.matrix[0][1] == 0) continue; if (!condit(tmp.matrix[0][1]) || !judge(0, 1) || !v[trans(0, 1)]) continue; tmp.matrix[4][1] = sum - sumallr[4]; if (!condit(tmp.matrix[4][1]) || !judge(4, 1) || !v[trans(4, 1)]) continue; tmp.matrix[2][1] = sum - sumallc[1] - tmp.matrix[0][1] - tmp.matrix[4][1]; if (!condit(tmp.matrix[2][1]) || !judge(2, 1) || !v[trans(1, 2)]) continue; tmp.matrix[1][0] = sum - sumallr[1] - tmp.matrix[1][2]; if (tmp.matrix[1][0] == 0) continue; if (!condit(tmp.matrix[1][0]) || !judge(1, 0) || !v[trans(1, 1)]) continue; tmp.matrix[2][0] = sum - sumallr[2] - tmp.matrix[2][1]; if (tmp.matrix[2][0] == 0) continue; if (!condit(tmp.matrix[2][0]) || !judge(2, 0) || !v[trans(2, 1)]) continue; if (sumallc[0] + tmp.matrix[1][0] + tmp.matrix[2][0] != sumallr[3]) continue; tmp.matrix[3][0] = sum - sumallr[3]; if (tmp.matrix[3][0] == 0) continue; if (!condit(tmp.matrix[3][0]) || !v[trans(0, 2)] || !v[trans(3, 1)]) continue; ans[anslen++] = tmp; } adds(3, 2, -1); } adds(4, 2, -1); } adds(0, 3, -1); adds(2, 3, -1); } adds(4, 3, -1); } adds(3, 4, -1); adds(2, 4, -1); } adds(1, 4, -1); } adds(4, 0, -1); adds(3, 1, -1); } adds(1, 3, -1); } adds(0, 4, -1); } } sort(ans, ans + anslen); for (int k = 0; k < anslen; ++k) { if (k != 0) printf("\n"); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 5; ++j) printf("%d", ans[k].matrix[i][j]); printf("\n"); } } return 0; }
