HDU-4326-game-probability DP

The answer to this question is very simple. N people are given each time, 4 people are playing the game, others are waiting, the winner continues, and the loser is ranked at the end. Those who win or win m times in a row become the final winner, what is the probability that the k-th person will eventually win?

For this question, we should first establish a model like this: X1 won the I game first, and is playing the game on X2, X3, and X4, followed by X5, X6 ,......, XN wait. P [I] [J] indicates the probability that the current XJ wins when X1 wins the I first.

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4326
I <m

P [I] [J] = P [I + 1] [J] * 1/4 + P [1] [n-2] * 3/4 j = 1

P [I + 1] [n-2] * 1/4 + P [1] [1] * 1/4 + P [1] [n-1] * 2/4 J = 2

P [I + 1] [n-1] * 1/4 + P [1] [1] * 1/4 + P [1] [n-1] * 1/4 + P [1] [N] * 1/4 J = 3

P [I + 1] [N] * 1/4 + P [1] [1] * 1/4 + P [1] [N] * 2/4 J = 4

P [I + 1] [J-3] * 1/4 + P [1] [J-3] * 3/4 j> = 4

I = m

P [I] [J] = 1 j = 1

0 J! = 1

Use the fractional form Gaussian elimination element to solve the equation.

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>using namespace std;#define maxn 102#define eps 1e-10double g[maxn][maxn];double x[maxn];int n,m,k;void add(int cnt,int i,int j,double val){    int t=i*n+j;    if(i==m){        if(j==1)    g[cnt][n*m+1]+=-1.0*val;        return;    }    g[cnt][t]+=val;}void gauss(int n,int m){    int row,col,i,j,k;    for(row=1,col=1;row<n,col<m;row++,col++){        k=row;        for(i=row+1;i<=n;i++)            if(fabs(g[i][col])>fabs(g[k][col])) k=i;        if(k!=row){            for(i=col;i<=m;i++) swap(g[k][i],g[row][i]);        }        if(fabs(g[row][col])<eps)   continue;        for(i=row+1;i<=n;i++){            if(fabs(g[i][col])<eps) continue;            double t=g[i][col]/g[row][col];            g[i][col]=0.0;            for(j=col+1;j<=m;j++)   g[i][j]-=t*g[row][j];        }    }    for(i=n;i>=1;i--){        x[i]=g[i][m];        for(j=i+1;j<=n;j++)    x[i]-=x[j]*g[i][j];        x[i]/=g[i][i];    }}int main(){    int i,j,cs,nn=0;    scanf("%d",&cs);    while(cs--){        scanf("%d%d%d",&n,&m,&k);        memset(g,0,sizeof(g));        int cnt=0;        for(i=0;i<m;i++)            for(j=1;j<=n;j++){                cnt++;                add(cnt,i,j,1.0);                if(j==1){                    add(cnt,i+1,j,-0.25);                    add(cnt,1,n-2,-0.75);                }                else if(j==2){                    add(cnt,i+1,n-2,-0.25);                    add(cnt,1,1,-0.25);                    add(cnt,1,n-1,-0.5);                }                else if(j==3){                    add(cnt,i+1,n-1,-0.25);                    add(cnt,1,1,-0.25);                    add(cnt,1,n-1,-0.25);                    add(cnt,1,n,-0.25);                }                else if(j==4){                    add(cnt,i+1,n,-0.25);                    add(cnt,1,n,-0.5);                    add(cnt,1,1,-0.25);                }                else{                    add(cnt,i+1,j-3,-0.25);                    add(cnt,1,j-3,-0.75);                }            }        gauss(n*m,n*m+1);        printf("Case #%d: %.6lf\n",++nn,x[k]);    }    return 0;}