hdu 3111 DLX解数独

思路：裸的DLX解数独。关键是建图，感觉还不如写个dfs直接，DLX写这个的代码很烦。

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pb push_back
#define mp make_pair
#define Maxn 400010
#define Maxm 200010
#define LL __int64
#define Abs(x) ((x)>0?(x):(-x))
#define lson(x) (x<<1)
#define rson(x) (x<<1|1)
#define inf 100000
#define lowbit(x) (x&(-x))
#define clr(x,y) memset(x,y,sizeof(x))
#define Mod 1000000007
using namespace std;
int L[Maxn],R[Maxn],D[Maxn],U[Maxn],S[Maxn],C[Maxn],X[Maxn],Q[Maxn],H[],id;
int g[][];
void init(int m)
{
    int i;
    for(i=;i<=m;i++){
        D[i]=U[i]=i;
        L[i+]=i;
        R[i]=i+;
        S[i]=;
    }
    R[m]=;
    id=m+;
    clr(H,-);
}
void ins(int r,int c)
{
    D[id]=D[c];
    U[id]=c;
    U[D[c]]=id;
    D[c]=id;
    S[c]++;
    if(H[r]<)
        H[r]=L[id]=R[id]=id;
    else{
        L[id]=H[r];
        R[id]=R[H[r]];
        L[R[H[r]]]=id;
        R[H[r]]=id;
    }
    C[id]=c;
    X[id++]=r;
}
void Remove(int c)
{
    int i,j;
    R[L[c]]=R[c];
    L[R[c]]=L[c];
    for(i=D[c];i!=c;i=D[i]){
        for(j=R[i];j!=i;j=R[j]){
            D[U[j]]=D[j];
            U[D[j]]=U[j];
            S[C[j]]--;
        }
    }
}
void Resume(int c)
{
    int i,j;
    R[L[c]]=c;
    L[R[c]]=c;
    for(i=D[c];i!=c;i=D[i]){
        for(j=R[i];j!=i;j=R[j]){
            U[D[j]]=j;
            D[U[j]]=j;
            S[C[j]]++;
        }
    }
}
bool dfs(int k)
{
    int i,j,c,temp;
    if(R[]==){
        for(i=;i<k;i++){
            int r,k;
            temp=X[Q[i]];
            k=temp%;
            if(!k) k=;
            c=(temp%)/;
            if(temp%==) c=;
            if(temp%%) c++;
            r=temp/;
            if(temp%) r++;
            g[r][c]=k;
        }
        return true;
    }
    temp=inf;
    for(i=R[];i;i=R[i]){
        if(S[i]<temp){
            temp=S[i];
            c=i;
        }
    }
    Remove(c);
    for(i=D[c];i!=c;i=D[i]){
        Q[k]=i;
        for(j=R[i];j!=i;j=R[j])
            Remove(C[j]);
        if(dfs(k+))
            return true;
        for(j=L[i];j!=i;j=L[j])
            Resume(C[j]);
    }

    Resume(c);
    return false;
}
void build()
{
    int i,j,k,b,r,c;
    init();
    for(i=;i<=;i++){
        for(j=;j<=;j++){
            if(g[i][j]){
                r=(i-)*+(j-)*+g[i][j];
                c=(i-)*+g[i][j];
                ins(r,c);
                c=+(j-)*+g[i][j];
                ins(r,c);
                c=+(i-)*+j;
                ins(r,c);
                c=*+(((i-)/)*+(j+)/-)*+g[i][j];
                ins(r,c);
                continue;
            }
            for(k=;k<=;k++){
                r=(i-)*+(j-)*+k;
                c=(i-)*+k;
                ins(r,c);
                c=+(j-)*+k;
                ins(r,c);
                c=+(i-)*+j;
                ins(r,c);
                c=*+(((i-)/)*+(j+)/-)*+k;
                ins(r,c);
            }
        }
    }
}
int main()
{
    int t,i,j,f=;
    char str[];
    scanf("%d",&t);
    while(t--){
        clr(g,);
        for(i=;i<=;i++){
            scanf("%s",str);
            for(j=;j<;j++){
                if(str[j]!='?')
                    g[i][j+]=str[j]-'';
            }
        }
        if(t)
        scanf("%s",str);
        build();
        if(f)
            printf("---\n");
        f=;
        if(!dfs()){
            printf("impossible\n");
            continue;
        }
        for(i=;i<=;i++){
            for(j=;j<=;j++){
                printf("%d",g[i][j]);
            }
            printf("\n");
        }
    }
    return ;
}