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: sdj22251PROG: inflateLANG: 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+5using 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;}