HDU 3909 Sudoku（数独转DLX精确覆盖）

题目地址
题意：给你一个n，然后给你一个n^2阶的数独，n分别有2，3，4这3种情况，然后你判断这个数独是不是有唯一的一个解，如果没有解的话输出”No Solution”，多个解的话就是输出”Multiple Solutions”，否则判断这个唯一解是不是最小解（最小解就是把任何一个给定的数变为不确定的话就会有多组解）
思路：我们通过map去映射字符与数字的关系，然后我们在DLX中加一个变量记录有多少种情况，（剪枝：当大于等于2的时候说明一定是有多组解的所以我们在tot等于2的时候直接让他返回就好了，因为已经是多组解了，没有必要继续搜索下去），如果我们只有一组解的话，我们就枚举每一个不为’.’的点，把他置为’.’，再进行一边DLX判断他是不是有多组解，如果有一个没有的就是输出”Not Minimal” ，否则输出数独结果就好了。
PS：这题还是蛮考对数独转DLX精确覆盖的理解的，题目其实不难，只是需要细心。我就是忘记把我测试输出的东西删掉了导致找了一天的bug。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 110#define MAXSIZE 1010*1010#define LL __int64#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1 using namespace std;const LL mod = 1000000007;struct node {    int left, right, up, down, col, row;}mapp[MAXSIZE];int S[MAXSIZE], H[MAXSIZE];//S记录该列中1元素的个数int head, cnt;int ans[MAXSIZE];int len;int tot;struct Dancing_Links_X {    void init(int m) {        head = 0; tot = 0;        for (int i = 0; i <= m; i++) {            S[i] = 0;            mapp[i].up = mapp[i].down = i;            mapp[i].left = (i == 0 ? m : i - 1);            mapp[i].right = (i == m ? 0 : i + 1);        }        cnt = m;        memset(H, -1, sizeof(H));    }    void link(int x, int y) {        cnt++;        mapp[cnt].row = x;        mapp[cnt].col = y;        S[y]++;        mapp[cnt].up = mapp[y].up;        mapp[cnt].down = y;        mapp[mapp[y].up].down = cnt;        mapp[y].up = cnt;        if (H[x] == -1) H[x] = mapp[cnt].left = mapp[cnt].right = cnt;        else {            mapp[cnt].left = mapp[H[x]].left;            mapp[cnt].right = H[x];            mapp[mapp[H[x]].left].right = cnt;            mapp[H[x]].left = cnt;        }    }    void remove(int c) {//删去c这个点,以及关联的一列        mapp[mapp[c].right].left = mapp[c].left;        mapp[mapp[c].left].right = mapp[c].right;        for (int i = mapp[c].down; i != c; i = mapp[i].down) {            for (int j = mapp[i].right; j != i; j = mapp[j].right) {                mapp[mapp[j].down].up = mapp[j].up; mapp[mapp[j].up].down = mapp[j].down;                --S[mapp[j].col];            }        }    }    void resume(int c) {//恢复c这个点，以及关联的一列        for (int i = mapp[c].up; i != c; i = mapp[i].up) {            for (int j = mapp[i].left; j != i; j = mapp[j].left) {                ++S[mapp[j].col];                mapp[mapp[j].down].up = mapp[mapp[j].up].down = j;            }        }        mapp[mapp[c].right].left = mapp[mapp[c].left].right = c;    }    void dance1(int k) {        if (mapp[head].right == head) {            len = k;            tot++;            return;        }        int s = inf, c;        for (int t = mapp[head].right; t != head; t = mapp[t].right) {            if (S[t] < s) s = S[t], c = t;        }        remove(c);        for (int i = mapp[c].down; i != c; i = mapp[i].down) {            ans[k] = mapp[i].row;            for (int j = mapp[i].right; j != i; j = mapp[j].right) {                remove(mapp[j].col);            }            dance1(k + 1);            if (tot == 2) {                return;            }            for (int j = mapp[i].left; j != i; j = mapp[j].left) {                resume(mapp[j].col);            }        }        resume(c);    }    bool dance(int k) {        if (mapp[head].right == head) {            len = k;            //cout << endl << len << endl << endl;            return true;        }        int s = inf, c;        for (int t = mapp[head].right; t != head; t = mapp[t].right) {            if (S[t] < s) s = S[t], c = t;        }        remove(c);        for (int i = mapp[c].down; i != c; i = mapp[i].down) {            ans[k] = mapp[i].row;            for (int j = mapp[i].right; j != i; j = mapp[j].right) {                remove(mapp[j].col);            }            if (dance(k + 1)) {                return true;            }            for (int j = mapp[i].left; j != i; j = mapp[j].left) {                resume(mapp[j].col);            }        }        resume(c);        return false;    }}DLX;char mat[N][N];map<char, int>map1;map<int, char>map2;void build(int n, int m) {    DLX.init(m * m * 4);    for (int i = 0; i < m; i++) {        for (int j = 0; j < m; j++) {            if (mat[i][j] != '.') {                int r = (i * m + j) * m + map1[mat[i][j]];                int c1 = i * m + j + 1;                int c2 = m * m + i * m + map1[mat[i][j]];                int c3 = 2 * m * m + j * m + map1[mat[i][j]];                int c4 = 3 * m * m + (i / n * n + j / n) * m + map1[mat[i][j]];                DLX.link(r, c1);                DLX.link(r, c2);                DLX.link(r, c3);                DLX.link(r, c4);                S[c1] = S[c2] = S[c3] = S[c4] = -1;            }        }    }    for (int i = 0; i < m; i++) {        for (int j = 0; j < m; j++) {            if (mat[i][j] == '.') {                for (int k = 1; k <= m; k++) {                    int r = (i * m + j) * m + k;                    int c1 = i * m + j + 1;                    int c2 = m * m + i * m + k;                    int c3 = 2 * m * m + j * m + k;                    int c4 = 3 * m * m + (i / n * n + j / n) * m + k;                    if (~S[c1] && ~S[c2] && ~S[c3] && ~S[c4]) {                        DLX.link(r, c1);                        DLX.link(r, c2);                        DLX.link(r, c3);                        DLX.link(r, c4);                    }                }            }        }    }}int main() {    cin.sync_with_stdio(false);    int n, m;    map1.clear();    map1['1'] = 1;map1['2'] = 2;map1['3'] = 3;map1['4'] = 4;    map1['5'] = 5;map1['6'] = 6;map1['7'] = 7;map1['8'] = 8;    map1['9'] = 9;map1['A'] = 10;map1['B'] = 11;map1['C'] = 12;    map1['D'] = 13;map1['E'] = 14;map1['F'] = 15;map1['G'] = 16;    map2.clear();    map2[1] = '1';map2[2] = '2';map2[3] = '3';map2[4] = '4';    map2[5] = '5';map2[6] = '6';map2[7] = '7';map2[8] = '8';    map2[9] = '9';map2[10] = 'A';map2[11] = 'B';map2[12] = 'C';    map2[13] = 'D';map2[14] = 'E';map2[15] = 'F';map2[16] = 'G';    while (cin >> n) {        m = n*n;        for (int i = 0; i < m; i++) {            for (int j = 0; j < m; j++) {                cin >> mat[i][j];            }        }        build(n, m);        len = inf;        DLX.dance1(0);        if (tot == 0) {            cout << "No Solution" << endl;        }        else if (tot == 1) {            bool flag = true;            for (int i = 0; i < m; i++) {                for (int j = 0; j < m; j++) {                    if (mat[i][j] == '.') continue;                    char ch = mat[i][j];                    mat[i][j] = '.';                    build(n, m);                    len = inf;                    DLX.dance1(0);                    if (tot != 2) {                        flag = false;                        break;                    }                    mat[i][j] = ch;                }            }            if (!flag) {                cout << "Not Minimal" << endl;            }            else {                build(n, m);                len = inf;                DLX.dance(0);                for (int i = 0; i < len; i++) {                    int Orz = (ans[i] - 1) % m + 1;                    int x = (ans[i] - 1) / m / m;                    int y = (ans[i] - 1) / m % m;                    if (mat[x][y] != '.') continue;                    mat[x][y] = map2[Orz];                }                for (int i = 0; i < m; i++) {                    for (int j = 0; j < m; j++) {                        cout << mat[i][j];                    }                    cout << endl;                }            }        }        else {            cout << "Multiple Solutions" << endl;        }    }    return 0;}