您的位置:首页 > 其它

【HDOJ】3909 Sudoku

2015-11-18 20:51 375 查看
DLX的应用,基本题,注意maxnode开大点儿。

/* 3909 */
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

typedef struct DLX {
static const int maxc = 4*16*16+5;
static const int maxr = 16*16*16+5;
static const int maxnode = 16*16*16*5+5;

int n, sz;
int S[maxc];

int row[maxnode], col[maxnode];
int L[maxnode], R[maxnode], U[maxnode], D[maxnode];

int ansd, cnt, ans[maxr], ans_[maxr];

void init(int n_) {
cnt = 0;
n = n_;

rep(i, 0, n+1) {
L[i] = i-1;
R[i] = i+1;
U[i] = i;
D[i] = i;
}

L[0] = n;
R
= 0;

sz = n+1;
memset(S, 0, sizeof(S));
}

void addRow(int r, vi columns) {
int first = sz;
int size = SZ(columns);

rep(i, 0, size) {
int c = columns[i];

L[sz] = sz-1;
R[sz] = sz+1;

D[sz] = c;
U[sz] = U[c];
D[U[c]] = sz;
U[c] = sz;

row[sz] = r;
col[sz] = c;

++S[c];
++sz;
}

R[sz - 1] = first;
L[first] = sz - 1;
}

void remove(int c) {
L[R[c]] = L[c];
R[L[c]] = R[c];
for (int i=D[c]; i!=c; i=D[i]) {
for (int j=R[i]; j!=i; j=R[j]) {
U[D[j]] = U[j];
D[U[j]] = D[j];
--S[col[j]];
}
}
}

void restore(int c) {
L[R[c]] = c;
R[L[c]] = c;
for (int i=D[c]; i!=c; i=D[i]) {
for (int j=R[i]; j!=i; j=R[j]) {
U[D[j]] = j;
D[U[j]] = j;
++S[col[j]];
}
}
}

bool dfs(int d) {
if (R[0] == 0) {
ansd = d;
++cnt;
rep(i, 0, ansd)
ans_[i] = ans[i];
return cnt>1;
}

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]) {
ans[d] = row[i];
for (int j=R[i]; j!=i; j=R[j]) {
remove(col[j]);
}
if (dfs(d + 1))    return true;
for (int j=L[i]; j!=i; j=L[j]) {
restore(col[j]);
}
}
restore(c);

return false;
}

void solve(vi& v) {

dfs(0);

if (cnt == 1) {
v.clr();
rep(i, 0, ansd)
v.pb(ans_[i]);
}
}

} DLX;

DLX solver;
int n, n2, n2n2;
const int maxl = 20;
char M[maxl][maxl], M_[maxl][maxl];
const int SLOT     = 0;
const int ROW      = 1;
const int COL    = 2;
const int SUB    = 3;

int encode(int a, int b, int c) {
return a*n2n2 + b*n2 + c + 1;
}

void decode(int code, int& a, int& b, int& c) {
--code;
c = code % n2;
code /= n2;
b = code % n2;
code /= n2;
a = code;
}

int getVal(char c) {
if (c>='0' && c<='9')    return c-'1';
return c-'A'+9;
}

int getChar(int val) {
if (val < 9)    return val+'1';
return val-9+'A';
}

void init() {
n2 = n * n;
n2n2 = n2 * n2;
}

void solve(vi& ans) {
solver.init(4 * n2n2);

rep(r, 0, n2) {
rep(c, 0, n2) {
rep(v, 0, n2) {
if (M[r][c]=='.' || v==getVal(M[r][c])) {
vi columns;
columns.pb(encode(SLOT, r, c));
columns.pb(encode(ROW, r, v));
columns.pb(encode(COL, c, v));
columns.pb(encode(SUB, r/n*n+c/n, v));
solver.addRow(encode(r, c, v), columns);
}
}
}
}

solver.solve(ans);
}

int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif

vi ans, tmp;

while (scanf("%d", &n) != EOF) {
init();
rep(i, 0, n2)
scanf("%s", M[i]);

solve(ans);

if (solver.cnt == 0) {
puts("No Solution");
continue;
} else if (solver.cnt > 1) {
puts("Multiple Solutions");
continue;
}

// check if is minimal
bool flag = true;

rep(r, 0, n2) {
rep(c, 0, n2) {
if (M[r][c] != '.') {
char ch = M[r][c];
M[r][c] = '.';
solve(tmp);
M[r][c] = ch;
if (solver.cnt <= 1) {
flag = false;
goto _output;
}
}
}
}

_output:
if (flag) {
int sz = SZ(ans);
rep(i, 0, sz) {
int r, c, v;
decode(ans[i], r, c, v);
char ch = getChar(v);
M[r][c] = ch;
}

rep(i, 0, n2)
puts(M[i]);
} else {
puts("Not Minimal");
}
}

#ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: