Light OJ 1018 status compression DP

Http://www.lightoj.com/volume_showproblem.php? Problem = 1018.

Question: give you n coordinate points (n <= 16) and ask you how many lines can pass through all points

Practice:

1: N is very small. It should be visually compressed DP

2: a single line will go through at least two points (otherwise it will be too deficient), so we can enumerate two points, preprocessing the state of the point set passing through the straight line of the two points (DP [I] [J])

3: Then you can perform the Memory search. Each time you enumerate the straight lines of two vertices in the S state, all vertices in the straight line are removed.

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int inf = ~0u>>2;int dp[20][20];int n;int f[1<<16];struct point {    int x,y;}in[20];int cross(point a,point b,point c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}bool col(int a,int b,int c){    return cross(in[a],in[b],in[c])==0;}void init(){    memset(dp,0,sizeof(dp));    for(int i=0;i<n;i++){        for(int j=i+1;j<n;j++){            for(int k=0;k<n;k++){                if(col(i,j,k)){                    dp[i][j]+=(1<<k);                }            }        }    }}int DP(int s){    if(f[s]!=inf) return f[s];    int cnt=0;    for(int i=0;i<n;i++) if(s&(1<<i)) cnt++;    if(cnt==0) return f[s]=0;    if(cnt<=2) return f[s]=1;    for(int i=0;i<n;i++)if(s&(1<<i))    {        for(int j=i+1;j<n;j++)if(s&(1<<j))        {            f[s]=min(f[s],DP(s-(s&dp[i][j]))+1);        }        break;//else TLE    }    return f[s];}int main(){    int t,ca=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d%d",&in[i].x,&in[i].y);        }        init();        fill(f,f+(1<<16),inf);        printf("Case %d: %d\n",ca++,DP((1<<n)-1));    }    return 0;}