HDU 2780 Su-Su-Sudoku（数独用DLX精确覆盖求解）

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题目地址
题意：给你一个数独残图，问你是不是能求出完整的图，完整的数独是什么样子的（唯一）。
思路：数独的限制条件就是4个，分别是

位置限制:每一格有且仅有一个数
列限制:每一列中每个数仅出现一次
行限制:每一行中每个数仅出现一次
区域限制:每个区域每个数仅出现一次

所以我们创建的矩阵要满足这些条件，所以我们划分出行的区间来表示这些限制条件

1-81行表示的是位置限制
82-162行表示的是列限制
163-243行表示的是行限制
244-324行表示的是区域限制

然后每一列表示的是该位置数的情况，比如mat[i][j]=2的话，那么（i*9+j）*9+2列的某些行（对应限制行）就为1，当如果是不确定的话（为没有填写的），那么就是把可以填的数字全部的列的某些行的值为1就好了，然后求一遍DLX的精确覆盖。
推荐知乎上一篇讲数独求解该怎样转化为精确覆盖问题：传送门

#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[N];int len;struct Dancing_Links_X {    void init(int m) {        head = 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;    }    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;int main() {    cin.sync_with_stdio(false);    int n, m;    char mat[N][N];    int T;    cin >> T;    while (T--) {        n = 9 * 9 * 9;//行        m = 9 * 9 * 4;//列        DLX.init(m);        for (int i = 0; i < 9; i++) {            for (int j = 0; j < 9; j++) {                cin >> mat[i][j];            }        }        for (int i = 0; i < 9; i++) {            for (int j = 0; j < 9; j++) {                if (mat[i][j] != '0') {                    int r = (i * 9 + j) * 9 + mat[i][j] - '0';                    int c1 = i * 9 + j + 1;                    int c2 = 81 + i * 9 + mat[i][j] - '0';                    int c3 = 162 + j * 9 + mat[i][j] - '0';                    int c4 = 243 + (i / 3 * 3 + j / 3) * 9 + mat[i][j] - '0';                    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 < 9; i++) {            for (int j = 0; j < 9; j++) {                if (mat[i][j] == '0') {                    for (int k = 1; k <= 9; k++) {                        int r = (i * 9 + j) * 9 + k;                        int c1 = i * 9 + j + 1;                        int c2 = 81 + i * 9 + k;                        int c3 = 162 + j * 9 + k;                        int c4 = 243 + (i / 3 * 3 + j / 3) * 9 + 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);                        }                    }                }            }        }        len = inf;        if (DLX.dance(0)) {            for (int i = 0; i < len; i++) {                int Orz = (ans[i] - 1) % 9 + 1;                int x = (ans[i] - 1) / 9 / 9;                int y = (ans[i] - 1) / 9 % 9;                mat[x][y] = Orz + '0';            }            for (int i = 0; i < 9; i++) {                for (int j = 0; j < 9; j++) {                    cout << mat[i][j];                }                cout << endl;            }        }        else {            cout << "Could not complete this grid." << endl;        }        if (T) {            cout << endl;        }    }    return 0;}

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