HDU 4305 Lightning (矩阵行列式求生成树个数+高斯逆元求行列式模板)

Lightning

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 999    Accepted Submission(s): 333

Problem Description

There are N robots standing on the ground (Don't know why. Don't know how). 

Suddenly the sky turns into gray, and lightning storm comes! Unfortunately, one of the robots is stuck by the lightning!

So it becomes overladen. Once a robot becomes overladen, it will spread lightning to the near one.


The spreading happens when: 
  Robot A is overladen but robot B not.
  The Distance between robot A and robot B is no longer than R.
  No other robots stand in a line between them.
In this condition, robot B becomes overladen. 

We assume that no two spreading happens at a same time and no two robots stand at a same position. 


The problem is: How many kind of lightning shape if all robots is overladen? The answer can be very large so we output the answer modulo 10007. If some of the robots cannot be overladen, just output -1. 

 

Input

There are several cases.
The first line is an integer T (T < = 20), indicate the test cases.
For each case, the first line contains integer N ( 1 < = N < = 300 ) and R ( 0 < = R < = 20000 ), indicate there stand N robots; following N lines, each contains two integers ( x, y ) ( -10000 < = x, y < = 10000 ), indicate the position of the robot. 

 

Output

One line for each case contains the answer.

 

Sample Input

33 2-1 00 11 03 2-1 00 01 03 1-1 00 11 0

 

Sample Output

31-1

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

 

Recommend

zhuyuanchen520

思路：题意比较明显，就是求一张无向图的生成树的个数。

如何用矩阵行列式求生成树的个数,这里有比较详细的讲解：http://wtommy.ycool.com/post.2218850.html

之后这题就是判断3点共线情况和模板的问题了。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define N 301#define MOD 10007bool g[N][N];struct node{    int x,y;} p[N];int n,R;inline bool ok(int i,int j){    if((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y)<=R*R) return 1;    return 0;}int aa[N][N];int gauss(int r, int c){    bool flag = false;    int coe = 1;    int i=0, t=0;    for(int j=0; j<c; ++j)    {        int index = i;        for(int k=i; k<r; ++k)        {            if(aa[k][j] > 0)            {                index = k;                break;            }        }        if(aa[index][j])        {            if(index != i)            {                for(int k=j; k<c; ++k)                {                    swap(aa[i][k],aa[index][k]);                }                flag = !flag;            }            for(int k=i+1; k<r; ++k)            {                if(aa[k][j])                {                    coe = (coe * aa[i][j]) % MOD;                    ++ t;                    for(int l=c-1; l>=j; --l)                    {                        aa[k][l] = (aa[k][l] * aa[i][j] - aa[i][l] * aa[k][j]) % MOD;                        if(aa[k][l] < 0)                        {                            aa[k][l] += MOD;                        }                    }                }            }            ++ i;        }    }    for(i=1;i<MOD;++i)    {        if((coe * i) % MOD == 1)        {            break;        }    }    int result = i;    for(i=0; i<r; ++i)    {        result = (result * aa[i][i]) % MOD;    }    if(flag)    {        result = MOD - result;    }    return result;}inline int Direction(int i,int j,int k){    return (p[j].x-p[i].x)*(p[k].y-p[i].y)-(p[k].x-p[i].x)*(p[j].y-p[i].y);}inline bool On_Segment(int i,int j,int k){    if(p[k].x>=min(p[i].x,p[j].x)&&p[k].x<=max(p[i].x,p[j].x)&&p[k].y>=min(p[i].y,p[j].y)&&p[k].y<=max(p[i].y,p[j].y))    return 1;    return 0;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&R);        memset(g,0,sizeof(g));        for(int i=1; i<=n; i++)            scanf("%d%d",&p[i].x,&p[i].y);        for(int i=1; i<=n; i++)            for(int j=i+1; j<=n; j++)                if(ok(i,j))                    g[i][j]=g[j][i]=1;            for(int k=1;k<=n;k++)            for(int i=1;i<=n;i++)            for(int j=i+1;j<=n;j++)            if(i!=k&&k!=j&&i!=j)            if(Direction(i,j,k)==0&&On_Segment(i,j,k))            g[i][j]=g[j][i]=0;            memset(aa,0,sizeof(aa));            for(int i=1;i<=n;i++)            for(int j=i+1;j<=n;j++)            if(g[i][j])            {                 aa[i-1][i-1]++;                 aa[j-1][j-1]++;                 aa[i-1][j-1]=aa[j-1][i-1]=-1;            }            for(int i=0;i<n-1;i++)            for(int j=0;j<n-1;j++)            aa[i][j]=aa[i+1][j+1];            int ans=gauss(n-1,n-1);            if(ans==0) puts("-1");            else                printf("%d\n",ans);    }    return 0;}