HDU 2828 Lamp

Problem Description

There are several switches and lamps in the room, however, the connections between them are very complicated. One lamp may be controlled by several switches, and one switch may controls at most two lamps. And what’s more, some connections are reversed by mistake, so it’s possible that some lamp is lighted when its corresponding switch is “OFF”! 

To make things easier, we number all the lamps from 1 to N, and all the switches 1 to M. For each lamps, we give a list of switches controlling it. For example, for Lamp 1, the list is “1 ON 3 OFF 9 ON”, that means Lamp 1 will be lighted if the Switch 1 is at the “ON” state OR the Switch 3 is “OFF” OR the Switch 9 is “ON”.

Now you are requested to turn on or off the switches to make all the lamps lighted. 

 

Input

There are several test cases in the input. The first line of each test case contains N and M (1 <= N,M <= 500), then N lines follow, each indicating one lamp. Each line begins with a number K, indicating the number of switches controlling this lamp, then K pairs of “x ON” or “x OFF” follow.

 

Output

Output one line for each test case, each contains M strings “ON” or “OFF”, indicating the corresponding state of the switches. For the solution may be not unique, any correct answer will be OK. If there are no solutions, output “-1” instead.

 

Sample Input

2 22 1 ON 2 ON1 1 OFF2 11 1 ON1 1 OFF

 

Sample Output

OFF ON-1
转换为01矩阵，dlx重复覆盖
#include<cstdio>#include<vector>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const ll maxn = 1005;int n, m, x, y, T;char s[maxn];vector<int> f[maxn][2];inline void read(int &ret){    char c;    do {        c = getchar();    } while (c < '0' || c > '9');    ret = c - '0';    while ((c = getchar()) >= '0' && c <= '9')        ret = ret * 10 + (c - '0');}struct DLX{    #define maxn 10005    #define F(i,A,s) for (int i=A[s];i!=s;i=A[i])    int L[maxn], R[maxn], U[maxn], D[maxn];    int row[maxn], col[maxn], ans[maxn], cnt[maxn];    int n, m, num, sz;    bool Flag;    void add(int now, int l, int r, int u, int d, int x, int y)    {        L[now] = l;    R[now] = r;    U[now] = u;        D[now] = d;   row[now] = x;  col[now] = y;    }    void reset(int n, int m)    {        Flag = false;        this->n = n;    this->m = m;        for (int i = 0; i <= m; i++)        {            add(i, i - 1, i + 1, i, i, 0, i);            cnt[i] = 0;        }        L[0] = m;     R[m] = 0;     sz = m + 1;    }    void insert(int x, int y)    {        int ft = sz - 1;        if (row[ft] != x)        {            add(sz, sz, sz, U[y], y, x, y);            U[D[sz]] = sz; D[U[sz]] = sz;        }        else        {            add(sz, ft, R[ft], U[y], y, x, y);            R[L[sz]] = sz; L[R[sz]] = sz;            U[D[sz]] = sz; D[U[sz]] = sz;        }        ++cnt[y];    ++sz;    }    //精确覆盖    void remove(int now)    {        R[L[now]] = R[now];        L[R[now]] = L[now];        F(i,D,now) F(j,R,i)        {            D[U[j]] = D[j];            U[D[j]] = U[j];            --cnt[col[j]];        }    }    void resume(int now)    {        F(i,U,now)    F(j,L,i)        {            D[U[j]] = j;            U[D[j]] = j;            ++cnt[col[j]];        }        R[L[now]] = now;        L[R[now]] = now;    }    int dfs(int x)    {        if (!R[0]) return 1;        int now = R[0];        F(i,R,0) if (cnt[now]>cnt[i]) now = i;        remove(now);        F(i,D,now)        {            ans[x] = row[i];            F(j,R,i) remove(col[j]);            if (dfs(x + 1)) return 1;            F(j,L,i) resume(col[j]);        }        resume(now);        return 0;    }    //精确覆盖    //重复覆盖    void Remove(int now)    {        F(i,D,now)         {            L[R[i]]=L[i];            R[L[i]]=R[i];        }    }    void Resume(int now)    {        F(i,U,now) L[R[i]]=R[L[i]]=i;    }    int vis[maxn];    int flag[maxn];    int A()    {        int dis=0;        F(i,R,0) vis[i]=0;        F(i,R,0) if (!vis[i])        {            dis++;    vis[i]=1;            F(j,D,i) F(k,R,j) vis[col[k]]=1;        }        return dis;    }    bool Dfs(int x)    {        if (!R[0]) {num=min(num,x); return true;}        else //if (x+A()<num)        {            int now=R[0];            F(i,R,0) if (cnt[now]>cnt[i]) now = i;            F(i,D,now)            if (!flag[row[i]^1])            {                ans[x]=row[i];                Remove(i);F(j,R,i) Remove(j);                flag[row[i]]=1;                if (Dfs(x+1)) return true;                flag[row[i]]=0;                F(j,L,i) Resume(j);Resume(i);            }        }        return false;    }    void display()    {        memset(vis,0,sizeof(vis));        for (int i=0;i<num;i++) vis[ans[i]>>1]=ans[i]&1;        for (int i=1;i+i<n;i++)         {            if (vis[i]) printf("ON"); else printf("OFF");             if (i+i==n-1) printf("\n"); else printf(" ");        }    }    void mul()    {        memset(flag,0,sizeof(flag));        num=0x7FFFFFFF;    }    //重复覆盖}dlx;int main(){    //read(T);    while (~scanf("%d%d",&n,&m))    {        dlx.reset(m+m+1,n);        for (int i=1;i<=m;i++)        {            f[i][0].clear();            f[i][1].clear();        }        for (int i=1;i<=n;i++)        {            scanf("%d",&x);            while (x--)            {                scanf("%d%s",&y,s);                if (s[1]=='N') f[y][0].push_back(i);                else f[y][1].push_back(i);            }        }        for (int i=1;i<=m;++i)            for (int j=0;j<2;++j)                for (int k=0;k<f[i][j].size();++k)                    dlx.insert(i+i-j+1,f[i][j][k]);        dlx.mul();        if (dlx.Dfs(0)) dlx.display();        else printf("-1\n");    }    return 0;}