HDU 4069 Squiggly Sudoku DLX

这是昨天周赛的题,我竟然不会怎么判断多解,后来一google,卧槽,我想复杂了。。。。。。直接看能搜出来几次就行了。

这题就是个变形,先floodfill一下,然后就是模板了

然后发现比大华的快了好几倍,然后加了输入挂后,瞬间成Best solution中的rank1了

/*
ID: sdj22251
PROG: inflate
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 9*9*9*9
#define INF 1000000000
#define N 9
#define M 9*9*9+5
using namespace std;
int L[MAXN], R[MAXN], C[MAXN], S[M], U[MAXN], D[MAXN], H[M], O[M], X[MAXN]; //X用来存储行,C用来存储列,H用来存储每行中的第一个结点,O用来存储结果
int cnt, head;
char mp[N + 5][N + 5], ans[N * N + 5];
bool vis[N * N * 4 + 5];
int ditu[N + 5][N + 5], pos[N * 5][N * 5];
int num, ct;
void link(int r, int c)
{
    S[c]++;
    C[cnt] = c;
    X[cnt] = r;
    U[cnt] = c;
    D[cnt] = D[c];
    U[D[c]] = cnt;
    D[c] = cnt;
    if(H[r] == -1)
    {
        H[r] = cnt;
        L[cnt] = R[cnt] = cnt;
    }
    else
    {
        L[cnt] = H[r];
        R[cnt] = R[H[r]];
        L[R[H[r]]] = cnt;
        R[H[r]] = cnt;
    }
    cnt++;
}
void init()
{
    cnt = 0;
    head = 0;
    num = 0;
    ct = 0;
    for(int i = 0; i <= N * N * 4; i++)
    {
        S[i] = 0;
        vis[i] = 0;
        D[i] = U[i] = i;
        R[i] = (i + 1) % (N * N * 4 + 1);
        L[i] = (i + N * N * 4) % (N * N * 4 + 1);
        cnt++;
    }
    memset(H, -1, sizeof(H));
}
void cal(int &r, int &cx, int &cy, int &ck, int &cg, int i, int j, int k)
{
    r = (i * N + j) * N + k - 1; //代表所属的行
    cg = i * N + j + 1;   //代表的是数独中i,j位置所属的列
    cx = N * N + i * N + k;  //代表数独中同一行所属的列
    cy = N * N * 2 + j * N + k; //代表数独中同一列所属的列
    ck = N * N * 3 + (pos[i][j] - 1) * N + k;//代表数独中同一宫所属的列
}
void search(int x, int y)
{
    pos[x][y] = ct;
    int ta = ditu[x][y];
    if(ta & 16) ta ^= 16;
    if(ta & 32) ta ^= 32;
    if(ta & 64) ta ^= 64;
    if(ta & 128) ta ^= 128;
    mp[x][y] = ta + '0';
    ta = ditu[x][y];
    if((ta & 16) == 0 && x >= 1 && pos[x - 1][y] == 0) search(x - 1, y);
    if((ta & 32) == 0 && y < N - 1 && pos[x][y + 1] == 0) search(x, y + 1);
    if((ta & 64) == 0 && x < N - 1 && pos[x + 1][y] == 0) search(x + 1, y);
    if((ta & 128) == 0 && y >= 1 && pos[x][y - 1] == 0) search(x, y - 1);
}
void readdata()
{
    int r, cx, cy, ck, cg;
    memset(pos, 0, sizeof(pos));
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
        scanf("%d", &ditu[i][j]);
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
        if(pos[i][j] == 0)
        {
            ++ct;
            search(i, j);
        }
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
            if(mp[i][j] != '0')
            {
                cal(r, cx, cy, ck, cg, i, j, mp[i][j] - '0');
                link(r, cx);
                link(r, cy);
                link(r, ck);
                link(r, cg);
                vis[cx] = vis[cy] = vis[ck] = vis[cg] = 1;
            }
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
            for(int k = 1; k <= N; k++)
            {
                cal(r, cx, cy, ck, cg, i, j, k);
                if(vis[cx] || vis[cy] || vis[ck] || vis[cg]) continue;
                link(r, cx);
                link(r, cy);
                link(r, ck);
                link(r, cg);
            }
}
void removes(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[C[j]]--;
        }
}
void resumes(int c)
{
    for(int i = U[c]; i != c; i = U[i])
        for(int j = L[i]; j != i; j = L[j])
        {
            U[D[j]] = j;
            D[U[j]] = j;
            S[C[j]]++;
        }
    L[R[c]] = c;
    R[L[c]] = c;
}
bool dfs(int k)
{
    if(R[head] == head)
    {
        num++;  //直接用num记录有多少个解
        for(int i = 0; i < k; i++)
            ans[X[O[i]] / N] = X[O[i]] % 9 + '1';
        if(num >= 2) return true;
        else return false;
    }
    int s = INF, c;
    for(int i = R[head]; i != head; i = R[i])
        if(s > S[i])
        {
            s = S[i];
            c = i;
        }
    removes(c);
    for(int i = U[c]; i != c; i = U[i])
    {
        O[k] = i;
        for(int j = R[i]; j != i; j = R[j])
            removes(C[j]);
        if(dfs(k + 1)) return true;
        for(int j = L[i]; j != i; j = L[j])
            resumes(C[j]);
    }
    resumes(c);
    return false;
}
int main()
{
    int T, cas = 0;
    scanf("%d", &T);
    while(T--)
    {
        init();
        readdata();
        dfs(0);
        printf("Case %d:\n", ++cas);
        if(num == 0) printf("No solution\n");
        else if(num == 1)
        {
            for(int i = 0; i < N * N; i++)
            {
                printf("%c", ans[i]);
                if((i + 1) % N == 0) printf("\n");
            }
        }
        else printf("Multiple Solutions\n");
    }
    return 0;
}


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