HDU_4770 Lights Against Dudely 状压+剪枝

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=4770

Lights Against Dudely

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2021    Accepted Submission(s): 595

Problem Description

Harry: "But Hagrid. How am I going to pay for all of this? I haven't any money." 
Hagrid: "Well there's your money, Harry! Gringotts, the wizard bank! Ain't no safer place. Not one. Except perhaps Hogwarts." 
— Rubeus Hagrid to Harry Potter. 
  Gringotts Wizarding Bank is the only bank of the wizarding world, and is owned and operated by goblins. It was created by a goblin called Gringott. Its main offices are located in the North Side of Diagon Alley in London, England. In addition to storing money and valuables for wizards and witches, one can go there to exchange Muggle money for wizarding money. The currency exchanged by Muggles is later returned to circulation in the Muggle world by goblins. According to Rubeus Hagrid, other than Hogwarts School of Witchcraft and Wizardry, Gringotts is the safest place in the wizarding world.
  The text above is quoted from Harry Potter Wiki. But now Gringotts Wizarding Bank is not safe anymore. The stupid Dudley, Harry Potter's cousin, just robbed the bank. Of course, uncle Vernon, the drill seller, is behind the curtain because he has the most advanced drills in the world. Dudley drove an invisible and soundless drilling machine into the bank, and stole all Harry Potter's wizarding money and Muggle money. Dumbledore couldn't stand with it. He ordered to put some magic lights in the bank rooms to detect Dudley's drilling machine. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)..... A 3×4 bank grid is shown below:

　　Some rooms are indestructible and some rooms are vulnerable. Dudely's machine can only pass the vulnerable rooms. So lights must be put to light up all vulnerable rooms. There are at most fifteen vulnerable rooms in the bank. You can at most put one light in one room. The light of the lights can penetrate the walls. If you put a light in room (x,y), it lights up three rooms: room (x,y), room (x-1,y) and room (x,y+1). Dumbledore has only one special light whose lighting direction can be turned by 0 degree,90 degrees, 180 degrees or 270 degrees. For example, if the special light is put in room (x,y) and its lighting direction is turned by 90 degrees, it will light up room (x,y), room (x,y+1 ) and room (x+1,y). Now please help Dumbledore to figure out at least how many lights he has to use to light up all vulnerable rooms.
　　Please pay attention that you can't light up any indestructible rooms, because the goblins there hate light. 

 

Input

　　There are several test cases.
　　In each test case:
　　The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 200).
　　Then a N×M matrix follows. Each element is a letter standing for a room. '#' means a indestructible room, and '.' means a vulnerable room. 
　　The input ends with N = 0 and M = 0

 

Output

　　For each test case, print the minimum number of lights which Dumbledore needs to put.
　　If there are no vulnerable rooms, print 0.
　　If Dumbledore has no way to light up all vulnerable rooms, print -1.

 

Sample Input

2 2####2 3#....#3 3####.####0 0

 

Sample Output

02-1

 

此题的关键在于点最多有15个，很容易想到使用状压。枚举装填与否的状态额，再枚举哪一个使用特殊的L，再枚举特殊L的姿态。但是，丧心病狂的出题人就会让你这么容易的过题吗！？于是漫长的剪枝就开始了：

1.首先预处理出每个点是否可以放一个普通的L，并用一个二进制状态can存下。每次先将特殊的姿态所在的位置加入can，这样每枚举出的状态i，如果i|can>can，说明这个i是没有意义的。

2.发现如果一个点的周围都是井号，那么就没有必要再找下去了，这个可以剪掉。

3.很给力的剪枝是先将二进制按含1的个数排序，再去寻找。

详见代码：

#include<iostream>#include<cstring>#include<vector>#include<algorithm>#include<cstdio>#define MAX_N 205#define MAX_M 205#define MAX_L 20using namespace std;char grid[MAX_N][MAX_M];bool used[MAX_N][MAX_M];int N,M;int tot=0,x[MAX_L],y[MAX_L];int dx[4]={1,1,-1,-1},dy[4]={1,-1,1,-1};int ddx[4]={0,1,0,-1},ddy[4]={1,0,-1,0};int can=0;struct node{    int one,value;};node two[16][1<<15];bool cmp(node a,node b){    return a.one<b.one;}int main(){    for(int i=0;i<16;i++)    {        for(int j=0;j<(1<<i);j++)        {            two[i][j].one=0;            for(int k=0;k<i;k++)                if((1<<k)&j)two[i][j].one++;            two[i][j].value=j;        }        sort(two[i],two[i]+(1<<i),cmp);    }    /*for(int i=0;i<5;i++,cout<<"------"<<endl)        for(int j=0;j<(1<<i);j++)        cout<<two[i][j].value<<" "<<two[i][j].one<<endl;*/    while(~scanf("%d%d",&N,&M))    {        int ans=100;        can=0;        if(N==0&&M==0)return 0;        memset(used,0,sizeof(used));        memset(x,0,sizeof(x));        memset(y,0,sizeof(y));        tot=0;        for(int i=0;i<N;i++)            for(int j=0;j<M;j++)            {                char c;                scanf(" %c",&c);                grid[i][j]=c;                if(c=='.')                    x[tot]=i,y[tot++]=j;            }        if(tot==0){printf("0\n");continue;}        bool fl=true;        for(int i=0;i<tot;i++)        {            int nearNum=0,daNum=0;            for(int j=0;j<4;j++)            {                int nnx=ddx[j]+x[i],nny=ddy[j]+y[i];                if(nnx>=0&&nnx<N&&nny>=0&&nny<M)                {                    nearNum++;                    daNum+=grid[nnx][nny]=='#';                }            }            //cout<<nearNum<<" "<<daNum<<endl;            if((nearNum==daNum)&&(nearNum>2)){printf("-1\n");fl=0;break;}            int nx=x[i]+dx[2],ny=y[i]+dy[2];            if((nx>=0&&nx<N&&grid[nx][y[i]]=='#')||(ny>=0&&ny<M&&grid[x[i]][ny]=='#'))can&=~(1<<i);            else                can|=(1<<i);        }        if(!fl)continue;        //cout<<can<<endl;        bool ffl=1;        for(int q=0;q<(1<<tot)&&ffl;q++)            for(int j=0;j<tot&&ffl;j++)            {                int i=two[tot][q].value;                int tmpCan=can|(1<<j);                if(!((1<<j)&i))continue;                if((tmpCan|i)>tmpCan)continue;                for(int k=0;k<4&&ffl;k++)                {                    memset(used,0,sizeof(used));                    int tmp=0,tmpAns=0;                    bool flag=true;                    for(int t=0;t<tot&&ffl;t++)                    {                        if(!((1<<t)&i))continue;                        int nx=x[t]+(t==j?dx[k]:dx[2]),ny=y[t]+(t==j?dy[k]:dy[2]);                        if(nx>=0&&nx<N&&grid[nx][y[t]]=='#'){flag=false;break;}                        if(ny>=0&&ny<M&&grid[x[t]][ny]=='#'){flag=false;break;}                         if(nx>=0&&nx<N){tmp+=!used[nx][y[t]];used[nx][y[t]]=1;}                        if(ny>=0&&ny<M){tmp+=!used[x[t]][ny];used[x[t]][ny]=1;}                        tmp+=!used[x[t]][y[t]];used[x[t]][y[t]]=1;                    }                    if(tmp==tot&&flag){ans=two[tot][q].one;ffl=0;break;}                }            }        //cout<<ans<<endl;        if(ans==100)            printf("-1\n");        else            printf("%d\n",ans);    }    return 0;}