SPOJNQUEEN-Yet Another N-Queen Problem

NQUEEN - Yet Another N-Queen Problem

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After solving Solution to the n Queens Puzzle by constructing, LoadingTime wants to solve a harder version of the N-Queen Problem. Some queens have been set on particular locations on the board in this problem. Can you help him??

Input

The input contains multiple test cases. Every line begins with an integer N (N<=50), then N integers followed, representing the column number of the queen in each rows. If the number is 0, it means no queen has been set on this row. You can assume there is at least one solution.

Output

For each test case, print a line consists of N numbers separated by spaces, representing the column number of the queen in each row. If there are more than one answer, print any one of them.

Example

Input:4 0 0 0 08 2 0 0 0 4 0 0 0Output:2 4 1 32 6 1 7 4 8 3 5

题意：n皇后问题，告诉你部分行皇后在哪，求出任意一种合法的摆放方法

解题思路：舞蹈链精确覆盖，不过它只需要保证行和列都被覆盖到了就好，斜线不需要所有的斜线都被覆盖到

#include <iostream>    #include <cstdio>    #include <cstring>    #include <string>    #include <algorithm>    #include <cctype>    #include <map>    #include <cmath>    #include <set>    #include <stack>    #include <queue>    #include <vector>    #include <bitset>    #include <functional>    using namespace std;#define LL long long    const int INF = 0x3f3f3f3f;const int maxn = 500005;int n, m, x, y, tot;int p[10900][3], vis[maxn], a[100];struct DLX{int L[maxn], R[maxn], U[maxn], D[maxn];int row[maxn], col[maxn], sum[maxn], ans[maxn];int n, m, num, cnt;void add(int k, int l, int r, int u, int d, int x, int y){L[k] = l;   R[k] = r;   U[k] = u;D[k] = d;   row[k] = x;  col[k] = y;}void reset(int n, int m){num = 0x7FFFFFF;this->n = n;   this->m = m;for (int i = 0; i <= m; i++){add(i, i - 1, i + 1, i, i, 0, i);sum[i] = 0;}L[0] = m, R[m] = 0, cnt = m + 1;}void insert(int x, int y){int temp = cnt - 1;if (row[temp] != x){add(cnt, cnt, cnt, U[y], y, x, y);U[D[cnt]] = cnt; D[U[cnt]] = cnt;}else{add(cnt, temp, R[temp], U[y], y, x, y);R[L[cnt]] = cnt; L[R[cnt]] = cnt;U[D[cnt]] = cnt; D[U[cnt]] = cnt;}sum[y]++, cnt++;}void remove(int k){R[L[k]] = R[k], L[R[k]] = L[k];for (int i = D[k]; i != k; i = D[i])for (int j = R[i]; j != i; j = R[j]){D[U[j]] = D[j];U[D[j]] = U[j];sum[col[j]]--;}}void resume(int k){R[L[k]] = k, L[R[k]] = k;for (int i = D[k]; i != k; i = D[i])for (int j = R[i]; j != i; j = R[j]){D[U[j]] = j;U[D[j]] = j;sum[col[j]]++;}}bool dfs(int k, int p){if (k == p) { num = min(k, num); return true; }if (!R[0]) return false;int now = R[0];for (int i = now; i <= p; i = R[i])if (sum[now] > sum[i]) now = i;remove(now);for (int i = D[now]; i != now; i = D[i]){ans[k] = row[i];for (int j = R[i]; j != i; j = R[j]) remove(col[j]);if (dfs(k + 1, p)) return true;for (int j = L[i]; j != i; j = L[j]) resume(col[j]);}resume(now);return false;}void display(int k){for (int i = 0; i < num; i++) a[p[ans[i]][0]] = p[ans[i]][1];for (int i = 1; i <= k; i++) printf("%d%s", a[i], i == k ? "\n" : " ");}}dlx;void insert(int x, int y, int z){int a = x, b = n + y, c = n + n + x + y - 1, d = 4 * n - 1 + (n - x) + y;if (vis[a] || vis[b] || vis[c] || vis[d]) return;tot++;p[tot][0] = x, p[tot][1] = y;dlx.insert(tot, a);dlx.insert(tot, b);dlx.insert(tot, c);dlx.insert(tot, d);if (z) vis[a] = vis[b] = vis[c] = vis[d] = 1;}int main(){while (~scanf("%d", &n)){tot = 0;dlx.reset(n * n, n * 6 - 2);memset(vis, 0, sizeof vis);for (int i = 1; i <= n; i++){scanf("%d", &x);if (x) insert(i, x, 1);}for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) insert(i, j, 0);if (dlx.dfs(0, n)) dlx.display(n);}return 0;}