HDU2828--Lamp(Dancing Links)

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
思路：要满足灯全亮DLX重复覆盖就可以搞定。但是每个开关只可以有一种状态。所以多个vis数组标记。
#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 MaxM = 2080;const int MaxN = 2080;const int maxnode = MaxN * MaxM;const int INF = 0x3f3f3f3f;#define eps 1e-8int N,M;bool vis[2080];struct DLX  {      int n,m,size;      int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];      int H[MaxN],S[MaxM];      int ansd;      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-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;          }      }      void remove(int c)      {          for(int i = D[c];i != c;i = D[i])              L[R[i]] = L[i], R[L[i]] = R[i];      }      void resume(int c)      {          for(int i = U[c];i != c;i = U[i])              L[R[i]] = R[L[i]] = i;      }          bool Dance(int d)      {          if(R[0] == 0)          {              return 1;          }          int c = R[0];          for(int i = R[0];i != 0;i = R[i])              if(S[i] < S[c])                  c = i;          for(int i = D[c];i != c;i = D[i])          {  if((Row[i]&1) && (vis[Row[i]+1]))continue;if(((Row[i]&1) == 0) && vis[Row[i]-1])continue;            remove(i);  vis[Row[i]] = 1;            for(int j = R[i];j != i;j = R[j])remove(j);              if(Dance(d+1))return 1;            for(int j = L[i];j != i;j = L[j])resume(j);              resume(i);  vis[Row[i]] = 0;        }return 0;    }  };  DLX g;  int main(){//freopen("in.txt","r",stdin);while(scanf("%d%d",&N,&M)==2){memset(vis,0,sizeof(vis));g.init(2*M,N);for(int i = 1;i <= N;i++){int k;scanf("%d",&k);for(int j = 1;j <= k;j++){int x;scanf("%d",&x);char ope[4];scanf("%s",ope);if(ope[1] == 'N')g.Link(2*x-1,i);else g.Link(2*x,i);}}if(!g.Dance(0)){printf("-1\n");continue;}else {for(int i = 1;i <= M;i++){if(vis[i<<1])cout << "OFF";else cout << "ON";if(i == M)cout << endl;else cout << " ";}}}}

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