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POJ 2676 Sudoku (DLX解数独)

2015-10-02 10:50 369 查看
题目大意:

经典数独, 9*9

每行每列每个九宫格都有1~9, 多解输出任意一解

大致思路:

没什么好说的数独了....和POJ 3074一样..

代码如下:

Result  :  Accepted     Memory  :  216 KB     Time  :  0 ms

/*
* Author: Gatevin
* Created Time: 2015/10/2 8:52:27
* File Name: Sakura_Chiyo.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

char in[800];

struct DLX
{
#define maxn 9*9*9 + 10
#define maxm 9*9*4 + 10
#define maxnode 9*9*9*4 + 100
int n, m, size;
int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode];
int H[maxn], S[maxm];
int ansd, ans[maxn];
struct State
{
int x, y, val;
State(int _x, int _y, int _val)
{
x = _x, y = _y, val = _val;
}
State(){}
};
State s[maxn];

void init(int _n, int _m)
{
n = _n;
m = _m;
for(int i = 0; i <= m; i++)
{
S[i] = 0;
U[i] = D[i] = i;
L[i] = i - 1;
R[i] = i + 1;
}
R[m] = 0; L[0] = m;
size = m;
for(int i = 1; i <= n; i++) H[i] = -1;
}
void Link(int r, int c)
{
++S[Col[++size] = c];
Row[size] = r;
D[size] = D[c];
U[D[c]] = size;
U[size] = c;
D[c] = size;
if(H[r] < 0) H[r] = L[size] = R[size] = size;
else
{
R[size] = R[H[r]];
L[R[H[r]]] = size;
L[size] = H[r];
R[H[r]] = size;
}
}
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 resume(int c)
{
for(int i = U[c]; i != c; i = U[i])
for(int j = L[i]; j != i; j = L[j])
++S[Col[U[D[j]] = D[U[j]] = j]];
L[R[c]] = R[L[c]] = c;
}
bool Dance(int dep)
{
if(R[0] == 0)
{
ansd = dep;
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])
{
ans[dep] = Row[i];
for(int j = R[i]; j != i; j = R[j]) remove(Col[j]);
if(Dance(dep + 1)) return true;
for(int j = L[i]; j != i; j = L[j]) resume(Col[j]);
}
resume(c);
return false;
}
void solve()
{
int tn = 0, tm = 9*9*4;
int len = strlen(in);
for(int i = 0; i < len; i++)
if(in[i] == '0') tn += 9;
else tn++;
init(tn, tm);
tn = 0;
for(int i = 0; i < 9; i++)
for(int j = 0; j < 9; j++)
{
char c = in[i*9 + j];
if(c == '0')
{
for(int k = 1; k <= 9; k++)
{
s[++tn] = State(i, j, k);
Link(tn, i*9 + k);
Link(tn, j*9 + k + 9*9);
Link(tn, (i/3 * 3 + j / 3)*9 + k + 9*9*2);
Link(tn, i*9 + j + 1 + 9*9*3);
}
}
else
{
s[++tn] = State(i, j, c - '0');
Link(tn, i*9 + c - '0');
Link(tn, j*9 + c - '0' + 9*9);
Link(tn, (i/3 * 3 + j / 3)*9 + c - '0' + 9*9*2);
Link(tn, i*9 + j + 1 + 9*9*3);
}
}
if(!Dance(0)) puts("No answer");
else
{
for(int i = ansd - 1; i >= 0; i--)
{
int sel = ans[i];
in[s[sel].x*9 + s[sel].y] = s[sel].val + '0';
}
for(int i = 0; i < 9*9; i++)
{
putchar(in[i]);
if(i % 9 == 8) putchar('\n');
}
}
}
};

DLX dlx;

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
for(int i = 0; i < 9; i++)
scanf("%s", in + i*9);
dlx.solve();
}
return 0;
}
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标签:  POJ 2676 Sudoku DLX