UVALive 2659 数独 DLX模板

建图：

　　从1到16枚举所有的行、列上放的数。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <cctype>
#include <time.h>

using namespace std;

const int INF = 1<<30;
const int MAXN = 1055;
const int MAXNODE = 16555;
const int MAXR = 1055;

struct DLX {
// 行编号从1开始，列编号为1~n，结点0是表头结点；结点1~n是各列顶部的虚拟结点
    int n, sz; // 列数，结点总数
    int S[MAXN]; //各列结点数

    int row[MAXNODE], col[MAXNODE]; //各结点行列编号
    int L[MAXNODE], R[MAXNODE], U[MAXNODE], D[MAXNODE]; //十字链表

    int ansd, ans[MAXR]; // 解

    void init(int n)  { //n是列数
        this->n = n;

        //虚拟结点
        for (int i = 0; i <= n; i++) {
            U[i] = i; D[i] = i; L[i] = i-1; R[i] = i+1;
        }
        R[n] = 0; L[0] = n;
        sz = n+1;
        memset(S, 0, sizeof(S));
    }

    void addRow(int r, vector<int> columns) {
        //这一行的第一个结点
        //行没有设头结点，每一行连成一个环形
        int first = sz;
        for (int i = 0; i < columns.size(); i++) {
            int c = columns[i];
            L[sz] = sz-1; R[sz] = sz+1; D[sz] = c; U[sz] = U[c];
            D[U[c]] = sz; U[c] = sz;
            row[sz] = r; col[sz] = c;
            S[c]++; sz++;
        }
        R[sz-1] = first; L[first] = sz-1;
    }

    //顺着链表A，遍历s外的其他元素
    #define FOR(i, A, s) for (int i = A[s]; i != s; i = A[i])

    void remove(int c) { //删除c列
        L[R[c]] = L[c]; R[L[c]] = R[c];
        for (int i = D[c]; i != c; i = D[i]) // 对于每一个c列不为0的所有行
            for (int j = R[i]; j != i; j = R[j]) { //删除这一整行
                U[D[j]] = U[j]; D[U[j]] = D[j]; S[col[j]]--;
            }
    }

    void restore(int c) { //回连c列
        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[col[j]]++;
            }
        L[R[c]] = c; R[L[c]] = c;
    }

    bool dfs(int d) { //d为递归深度
        if (R[0] == 0) { //找到解
            ansd = d; //记录解的长度
            return true;
        }

        //找S最小的C列
        int c = R[0]; //第一个未删除的列
        for (int i = R[0]; i != 0; i = R[i]) if (S[i]<S[c]) c = i;

        remove(c); //删除第c列
        for (int i = D[c]; i != c; i = D[i]) { //用结点i所在的行覆盖第c列
            ans[d] = row[i];
            for (int j = R[i]; j != i; j = R[j]) remove(col[j]); //删除节结点i所在行覆盖第c列
            if (dfs(d+1)) return true;
            for (int j = L[i]; j != i; j = L[j]) restore(col[j]); // 恢复
        }
        restore(c); //恢复
        return false;
    }

    bool solve(vector<int> &v) {
        v.clear();
        if(!dfs(0)) return false;
        for (int i = 0; i < ansd; i++) v.push_back(ans[i]);
        return true;
    }
};

const int SLOT = 0;
const int ROW = 1;
const int COL = 2;
const int SUB = 3;

inline int encode(int a, int b, int c) {
    return a*256 + b*16 + c + 1;
}

inline void decode(int code, int &a, int &b, int &c) {
    code--;
    c = code%16; code /= 16;
    b = code%16; code /= 16;
    a = code;
}

char puzzle[16][20];

bool read() {
    for (int i = 0; i < 16; i++)
        if (scanf("%s", puzzle[i])!=1) return false;
    return true;
}

DLX solver;

int main() {
    #ifdef Phantom01
        freopen("LA2659.txt", "r", stdin);
    #endif //Phantom01

    int T= 0;

    while (read()) {
        if (T!=0) puts(""); T++;
        solver.init(1024);
        for (int r = 0; r < 16; r++)
            for (int c = 0; c < 16; c++)
                for (int v = 0; v < 16; v++)
                    if (puzzle[r][c]=='-'|| puzzle[r][c]=='A'+v) {
                        vector<int> columns;
                        columns.push_back(encode(SLOT, r, c));
                        columns.push_back(encode(ROW, r, v));
                        columns.push_back(encode(COL, c, v));
                        columns.push_back(encode(SUB, (r/4)*4+c/4, v));
                        solver.addRow(encode(r, c, v), columns);
                    }
        vector<int> ans;
        if (!solver.solve(ans)) while (1) ;

        for (int i = 0; i < ans.size(); i++) {
            int r, c, v;
            decode(ans[i], r, c, v);
            puzzle[r][c] = 'A'+v;
        }
        for (int i = 0; i < 16; i++) {
            printf("%s
", puzzle[i]);
        }
    }

    return 0;
}