[DLX Repeat overlay] hdu 2828 Lamp

Test instructions

There are n lights m switches

The on and off states of each lamp can control whether a light is lit

Give n rows, which represent each lamp

Which state of the switch can make the first I light

Ideas:

Here's a question to be aware of.

If the on state of Switch 1 and the on state of Switch 2 can make the 1th light

Then the switch 1, 2 at the same time on the number 1th lamp is also bright. It means that as long as there is a switch to make the light light, the light is on.

Simple DLX Repeat Overlay

Behavior of two states per switch 2*m line, listed as n lights

At the same time as the search, mark which switch was used.

Then the other state is not going to work.

Code:

#include "stdio.h" #include "algorithm" #include "string.h" #include "iostream" #include "queue" #include "map" #include " Vector "#include" string "using namespace std; #define N 1005*1005#define RN 1005#define CN 1005int us[rn];struct dlx{int    N,m,c;    int u[n],d[n],l[n],r[n],row[n],col[n];    int H[RN],S[CN],CNT,ANS[RN];        void init (int _n,int _m) {n=_n;        M=_m;            for (int i=0; i<=m; i++) {s[i]=0;            U[i]=d[i]=i;            L[i]= (i==0?m:i-1);        R[i]= (i==m?0:i+1);        } c=m;    for (int i=1; i<=n; i++) h[i]=-1;        } void link (int x,int y) {C + +;        Row[c]=x;        Col[c]=y;        s[y]++;        U[c]=u[y];        D[c]=y;        D[u[y]]=c;        U[y]=c;        if (h[x]==-1) h[x]=l[c]=r[c]=c;            else {l[c]=l[h[x]];            R[C]=H[X];            R[l[h[x]]]=c;        L[h[x]]=c; }} void del (int x) {for (int i=d[x]; i!=x; I=d[i]) {r[l[i]]=R[i];        L[r[i]]=l[i];            }} void Rec (int x) {for (int i=u[x]; i!=x; I=u[i]) {r[l[i]]=i;        L[r[i]]=i;    }} int USED[CN];        int h () {int sum=0;        for (int i=r[0]; i!=0; i=r[i]) used[i]=0;                for (int i=r[0]; i!=0; I=r[i]) {if (used[i]==0) {sum++;                Used[i]=1;            For (int. j=d[i]; j!=i; j=d[j]) for (int k=r[j]; k!=j; k=r[k]) used[col[k]]=1;    }} return sum;        } int Dance (int x) {//if (X+h () >=cnt) return 0;            if (r[0]==0) {cnt=x;            Cnt=min (CNT,X);        return 1;        } int now=r[0];        for (int i=r[0]; i!=0; I=r[i]) {if (S[i]<s[now]) now=i; } for (int i=d[now]; i!=now; I=d[i]) {if (us[(row[i]-1)/2+1]==0) {us[(R                ow[i]-1)/2+1]=1;   Del (i);             Ans[x]=row[i];                for (int j=r[i]; j!=i; J=r[j]) del (j);                if (dance (x+1)) return 1;                for (int j=l[i]; j!=i; j=l[j]) rec (j);                Rec (i);            us[(row[i]-1)/2+1]=0;    }} return 0;    }} Dlx;int Main () {int n,m;        while (scanf ("%d%d", &n,&m)!=-1) {dlx.init (2*m,n);            for (int i=1; i<=n; i++) {int k;            scanf ("%d", &k);                while (k--) {int x;                Char y[21];                scanf ("%d%s", &x,y);                int tep=x*2-1;                if (strcmp (Y, "OFF") ==0) tep++;            Dlx.link (Tep,i);        }} memset (Us,0,sizeof (US));        int f=dlx.dance (0);        if (f==0) puts ("-1");            else {memset (us,0,sizeof (US));            for (int i=0;i<dlx.cnt;i++) {us[(dlx.ans[i]-1)/2+1]=dlx.ans[i]%2; } for (int i=1;i<=m;i++) printf (i==m? ") %s\n ":"%s ", us[i]==0?"        OFF ":" On "); }} return 0;}

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[DLX Repeat overlay] hdu 2828 Lamp