HDU4069(未AC)

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未AC代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const int maxnode = 100010;const int MaxM = 1010;const int MaxN = 1010;int anscnt=0;int mm[10][10];int ansmap[10][10];int markmap[10][10];int vis[10][10];struct DLX{    int n,m,size;    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];    //U D R L用来记录某个标号的节点的上下左右节点的标号    //Row Col用来记录某个标号的节点在矩阵中的行号和列号    int H[MaxN], S[MaxM];    //H是行头 S用来保存某一列中1的数量    int ansd, ans[MaxN];    void init(int _n,int _m)    {        n = _n;        m = _m;        //初始化列头        for(int i = 0;i <= m;i++)        {            S[i] = 0;            U[i] = D[i] = i;            L[i] = i-1;            R[i] = i+1;        }        R[m] = 0; L[0] = m;        size = m;        for(int i = 1;i <= n;i++)            H[i] = -1;    }    void Link(int r,int c)    {        ++S[Col[++size]=c];        Row[size] = r;        D[size] = D[c];        U[D[c]] = size;        U[size] = c;        D[c] = size;        if(H[r] < 0)H[r] = L[size] = R[size] = size;        else        {            R[size] = R[H[r]];            L[R[H[r]]] = size;            L[size] = H[r];            R[H[r]] = size;        }    }    //对某一列进行删除 同时删除列中为1的行    void remove(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[Col[j]];            }    }    //反着恢复状态    void resume(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]]=D[U[j]]=j;                ++S[Col[j]];            }        L[R[c]] = R[L[c]] = c;    }    //d为递归深度    void Dance(int d)    {        if(d>81) return;        if(anscnt>=2) return;        if(R[0] == 0)        {            anscnt++;            memcpy(ansmap,mm,sizeof(mm));            return ;        }        int c = R[0];        //一个优化  找到列中包含1最多的列 因为这样有助于减少递归深度 （很显然1多了 删掉的行也多 矩阵缩小得就快）        for(int i = R[0];i != 0;i = R[i])            if(S[i] < S[c])                c = i;        remove(c);        //搜索        for(int i = D[c];i != c;i = D[i])        {            mm[Row[i]][Col[i]]=(Row[i]-1)%9+1;            for(int j = R[i]; j != i;j = R[j])remove(Col[j]);            Dance(d+1);            for(int j = L[i]; j != i;j = L[j])resume(Col[j]);        }        resume(c);    }};DLX g;bool haswall(int x,int y, int d){    int tmp=mm[x][y]/16;    return tmp&(1<<d);}void dfs(int x,int y,int id){    if(vis[x][y]!=0)        return;    markmap[x][y]=id;    vis[x][y]=1;    if(!haswall(x, y, 0))    {        dfs(x-1,y,id);    }    if(!haswall(x, y, 1))    {        dfs(x,y+1,id);    }    if(!haswall(x, y, 2))    {        dfs(x+1,y,id);    }    if(!haswall(x, y, 3))    {        dfs(x,y-1,id);    }}int main(){    int cas;    scanf("%d",&cas);    int id=1;    for(int z=1;z<=cas;z++)    {        id=1;        anscnt=0;        memset(vis,0,sizeof(vis));        memset(markmap,0,sizeof(markmap));        for(int i=1;i<=9;i++)            for(int j=1;j<=9;j++)                scanf("%d",&mm[i][j]);        for(int i=1;i<=9;i++)            for(int j=1;j<=9;j++)            {                if(vis[i][j]==0)                    dfs(i,j,id++);            }        for(int i=1;i<=9;i++)            for(int j=1;j<=9;j++)                mm[i][j]%=16;        g.init(729,324);        for(int i=1;i<=9;i++)        {            for(int j=1;j<=9;j++)            {                int id1=(i-1)*9+j;                for(int k=1;k<=9;k++)                {                    if((!mm[i][j])||(mm[i][j]==k))                    {                        int _cx=(id1-1)*9+k;                        g.Link(_cx,id1);                        g.Link(_cx,81+(i-1)*9+k);                        g.Link(_cx,2*9*9+(j-1)*9+k);                        g.Link(_cx,3*9*9+(markmap[i][j]-1)*9+k);                    }                }            }        }        g.Dance(0);        printf("Case %d:\n",z);        if(anscnt==0)            printf("No Soltions\n");        if(anscnt==1)        {            for(int i=1;i<=9;i++)            {                for(int j=1;j<=9;j++)                {                    printf("%d",ansmap[i][j]);                }                printf("\n");            }        }        if(anscnt>1)            printf("Multiple Solutions\n");    }    return 0;}

金斌的ac代码

#include<cstdio>#include<cstdlib>#include<cstring>#include<vector>#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>using namespace std;const int N=9;const int mxn=N*N*N+10;const int mxm=4*N*N+10;const int mx=4*mxn+mxm;struct DLX{    int l[mx],r[mx],u[mx],d[mx];    int x[mx],y[mx];    int siz;    int ansd,ans[N*N+10],gans[N*N+10];    int h[mxn];    int s[mxm];    void init(int _n,int _m){        for(int i=0;i<=_m;i++){            u[i]=d[i]=i;            l[i]=i-1;            r[i]=i+1;            s[i]=0;        }        for(int i=1;i<=_n;i++){            h[i]=-1;        }        r[_m]=0;        l[0]=_m;        siz=_m;        ansd=0;    }    void link(int ro,int co){        //printf("%d %d\n",ro,co);        x[++siz]=ro;        y[siz]=co;        d[siz]=co;        u[siz]=u[co];        u[co]=siz;        d[u[siz]]=siz;        s[co]++;        if(h[ro]+1){            int hr=h[ro];            r[siz]=hr;            l[siz]=l[hr];            l[hr]=siz;            r[l[siz]]=siz;        }        else h[ro]=l[siz]=r[siz]=siz;    }    void removc(int c){        r[l[c]]=r[c];        l[r[c]]=l[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[y[j]]--;        }    }    void resumc(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]]=d[u[j]]=j;            s[y[j]]++;        }        r[l[c]]=l[r[c]]=c;    }    void dance(int dd){        if(dd>81){            return ;        }        if(ansd>=2)return ;        if(r[0]==0){            ansd++;            for(int i=1;i<=N*N;i++){                gans[i]=ans[i];            }            return ;        }        int c=r[0];        for(int i=r[0];i!=0;i=r[i]){            if(s[i]<s[c])c=i;        }        removc(c);        for(int i=d[c];i!=c;i=d[i]){            ans[(x[i]-1)/N+1]=(x[i]-1)%N+1;            for(int j=r[i];j!=i;j=r[j])removc(y[j]);            dance(dd+1);            for(int j=l[i];j!=i;j=l[j])resumc(y[j]);        }        resumc(c);    }}s;int mp[N+5][N+5];int id[N+5][N+5];int xx[5]={-1,0,1,0};int yy[5]={0,1,0,-1};int w[6]={16,32,64,128,256};int iid;void dfs(int x,int y,int iid){    id[x][y]=iid;    for(int i=0;i<4;i++){        int nx=x+xx[i];        int ny=y+yy[i];        if(nx>=1&&nx<=N&&ny>=1&&ny<=N&&(!id[nx][ny])){            int f=mp[x][y]%w[i+1];            if(f>=w[i])continue;            dfs(nx,ny,iid);        }    }}int main(void){    int t;    while(scanf("%d",&t)!=EOF){    for(int cas=1;cas<=t;cas++){        for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++)            scanf("%d",&mp[i][j]);        }        memset(id,0,sizeof(id));        iid=0;        for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++){                if(!id[i][j])                dfs(i,j,++iid);            }        }        for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++){                mp[i][j]%=16;            }        }        s.init(N*N*N,4*N*N);        for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++){                int id1=(i-1)*N+j;                for(int k=1;k<=N;k++){                    if((!mp[i][j])||(mp[i][j]==k)){                        int _cx=(id1-1)*N+k;                        s.link(_cx,id1);                        s.link(_cx,N*N+(i-1)*N+k);                        s.link(_cx,2*N*N+(j-1)*N+k);                        s.link(_cx,3*N*N+(id[i][j]-1)*N+k);                    }                }            }        }        printf("Case %d:\n",cas);        s.dance(0);        if(!s.ansd)puts("No solution");        else if(s.ansd>1)puts("Multiple Solutions");        else {            for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++){                printf("%d",s.gans[(i-1)*N+j]);            }            puts("");            }        }    }    }    return 0;}

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