DLX精确覆盖 poj2676 Sudoku
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传送门:点击打开链接
题意:求数独
思路:早就听说了DLX精确覆盖的强大,看了两天终于勉强算是看懂了,关于DLX的详细介绍可以参考下面3位大神总结的
DLX的原理:点击打开链接
DLX用C++的实现代码:点击打开链接
DLX的建图技巧:点击打开链接
要注意的几个地方就是,DLX有两种模式,一种是精确覆盖,一种是重复覆盖。
重复覆盖中包含了一个A*的估计函数,用来剪枝
然后就是把握好建图(感觉这个非常关键)
DLX中包含了许多链表的妙用,比如如何建立循环链表,如何循环除了本身以外的所有其他节点,如何在链表中删除和恢复删除等等,写的非常的飘逸~
#include<map>#include<set>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck printf("fuck")#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;const int MX = 1000 + 5;const int MN = 1000000 + 5;const int INF = 0x3f3f3f3f;int ans[MX][MX];struct DLX { int m, n; int H[MX], S[MX]; int Row[MN], Col[MN], rear; int L[MN], R[MN], U[MN], D[MN]; void Init(int _m, int _n) { m = _m; n = _n; rear = n; for(int i = 0; i <= n; i++) { S[i] = 0; L[i] = i - 1; R[i] = i + 1; U[i] = D[i] = i; } L[0] = n; R[n] = 0; for(int i = 1; i <= m; i++) { H[i] = -1; } } void Link(int r, int c) { int rt = ++rear; Row[rt] = r; Col[rt] = c; S[c]++; D[rt] = D[c]; U[D[c]] = rt; U[rt] = c; D[c] = rt; if(H[r] == -1) { H[r] = L[rt] = R[rt] = rt; } else { int id = H[r]; R[rt] = R[id]; L[R[id]] = rt; L[rt] = id; R[id] = rt; } } void Remove(int c) { R[L[c]] = R[c]; L[R[c]] = L[c]; for(int i = D[c]; i != c; i = D[i]) { for(int j = R[i]; j != i; j = R[j]) { D[U[j]] = D[j]; U[D[j]] = U[j]; S[Col[j]]--; } } } void Resume(int c) { for(int i = U[c]; i != c; i = U[i]) { for(int j = L[i]; j != i; j = L[j]) { D[U[j]] = U[D[j]] = j; S[Col[j]]++; } } R[L[c]] = L[R[c]] = c; } bool Dance(int cnt) { if(R[0] == 0) return true; int c = R[0]; for(int i = R[0]; i != 0; i = R[i]) { if(S[i] < S[c]) c = i; } Remove(c); for(int i = D[c]; i != c; i = D[i]) { for(int j = R[i]; j != i; j = R[j]) Remove(Col[j]); int r = Row[i]; ans[(r - 1) / 81 + 1][((r - 1) % 81) / 9 + 1] = ((r - 1) % 81) % 9 + 1; if(Dance(cnt + 1)) return true; for(int j = L[i]; j != i; j = L[j]) Resume(Col[j]); } Resume(c); return false; }} G;int S[MX][MX], vis[10];void check(int x, int y) { for(int i = 1; i <= 9; i++) vis[i] = 0; for(int i = 1; i <= 9; i++) { vis[S[i][y]] = vis[S[x][i]] = 1; } int tx = (x - 1) / 3 + 1, ty = (y - 1) / 3 + 1; for(int i = (tx - 1) * 3 + 1; i <= tx * 3; i++) { for(int j = (ty - 1) * 3 + 1; j <= ty * 3; j++) { vis[S[i][j]] = 1; } }}int main() { int T; //FIN; scanf("%d", &T); while(T--) { G.Init(9 * 9 * 9, 4 * 9 * 9); for(int i = 1; i <= 9; i++) { for(int j = 1; j <= 9; j++) { scanf("%1d", &S[i][j]); } } for(int i = 1; i <= 9; i++) { for(int j = 1; j <= 9; j++) { int tx = (i - 1) / 3 + 1, ty = (j - 1) / 3 + 1, tp = (tx - 1) * 3 + ty; if(!S[i][j]) { check(i, j); for(int k = 1; k <= 9; k++) { if(vis[k]) continue; int id = ((i - 1) * 9 + j - 1) * 9 + k; G.Link(id, (i - 1) * 9 + j); G.Link(id, 9 * 9 + (i - 1) * 9 + k); G.Link(id, 2 * 9 * 9 + (j - 1) * 9 + k); G.Link(id, 3 * 9 * 9 + (tp - 1) * 9 + k); } } else { int k = S[i][j]; int id = ((i - 1) * 9 + j - 1) * 9 + k; G.Link(id, (i - 1) * 9 + j); G.Link(id, 9 * 9 + (i - 1) * 9 + k); G.Link(id, 2 * 9 * 9 + (j - 1) * 9 + k); G.Link(id, 3 * 9 * 9 + (tp - 1) * 9 + k); } } } int ret = G.Dance(0); for(int i = 1; i <= 9; i++) { for(int j = 1; j <= 9; j++) { if(S[i][j]) printf("%d", S[i][j]); else printf("%d", ans[i][j]); } printf("\n"); } } return 0;}
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