【系列】高斯消元

bzoj 1013

题目大意：

给出$n$维球体上的$n+1$个点，求球心

思路：

设球心坐标$(x_1,x_2,x_3 cdots x_n)$

则对于任意两个点$(a_1,a_2 cdots a_n),(b_1,b_2 cdots b_n)$，得到$(x_1-a_1)^2+(x_2-a_2)^2 cdots (x_n-a_n)^2=(x_1-b_1)^2+(x_2-b_2)^2 cdots (x+n-b_n)^2$

移项后得到$2(a_1-b_1)x_1+2(a_2-b_2)x_2 cdots 2(a_n-b_n)x_n = {a_1}^2-{b_1}^2+{a_2}^2-{b_2}^2 cdots {a_n}^2-{b_n}^2$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
#define db double
#define inf 2139062143
#define MAXN 200100
#define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
#define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
#define ren for(register int i=fst[x];i;i=nxt[i])
#define Fill(x,t) memset(x,t,sizeof(x))
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-b+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;db a[15][15],g[15][15];
db sqr(db x) {return x*x;}
void Gauss()
{
    int pos;rep(i,0,n-1)
    {
        pos=n;rep(j,i,n-1)if(fabs(a[j][i])>fabs(a[pos][i])) pos=j;
        if(pos!=i) rep(j,0,n) swap(a[i][j],a[pos][j]);
        dwn(j,n,i) rep(k,i+1,n-1) a[k][j]-=a[k][i]/a[i][i]*a[i][j];
    }
    dwn(i,n-1,0) {rep(j,i+1,n-1) a[i][n]-=a[j][n]*a[i][j];a[i][n]/=a[i][i];}
}
int main()
{
    n=read();rep(i,0,n) rep(j,0,n-1) scanf("%lf",&g[i][j]);
    rep(i,1,n) rep(j,0,n-1)
        a[i-1][j]=2*(g[i][j]-g[i-1][j]),a[i-1][n]+=sqr(g[i][j])-sqr(g[i-1][j]);
    Gauss();rep(i,0,n-1) printf("%.3lf ",a[i][n]);
}

bzoj 3270

题目大意：

一开始两个人在$s,t$两个点，当一个人在$i$点时，有$p_i$的概率不动，有$1-p_i$的概率向相邻的点等概率移动

如果两个人在一个点相遇就停止，求他们在每个点相遇的概率

思路：

设$f(i,j)$表示两个人在$i,j$的概率，设$x,y$分别为$i,j$的前驱

很容易得到方程$f(i,j)=sum (i==x?p_x:frac{1-p_x}{d_x})*(j==y?p_y:frac{1-p_y}{d_y})$

对于$f(s,t)$概率为$1$，其余为$0$，一共$n^2$个方程消元即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
#define db double
#define inf 2139062143
#define MAXN 200100
#define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
#define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
#define ren for(register int i=fst[x];i;i=nxt[i])
#define Fill(x,t) memset(x,t,sizeof(x))
#define pls(a,b) (a+b)%MOD
#define mns(a,b) (a-b+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,ss,tt,mp[35][35],d[35];db a[410][410],p[35];
void Gauss(int n)
{
    int pos;rep(i,0,n-1)
    {
        pos=n;rep(j,i,n-1)if(fabs(a[j][i])>fabs(a[pos][i])) pos=j;
        if(pos!=i) rep(j,0,n) swap(a[i][j],a[pos][j]);
        dwn(j,n,i) rep(k,i+1,n-1) a[k][j]-=a[k][i]/a[i][i]*a[i][j];
    }
    dwn(i,n-1,0) {rep(j,i+1,n-1) a[i][n]-=a[j][n]*a[i][j];a[i][n]/=a[i][i];}
}
inline int t(int x,int y) {return (x-1)*n+y-1;}
int main()
{
    n=read(),m=read(),ss=read(),tt=read();int x,y;
    while(m--) x=read(),y=read(),mp[x][y]=mp[y][x]=1,d[x]++,d[y]++;
    rep(i,1,n) scanf("%lf",&p[i]),mp[i][i]=1;a[t(ss,tt)][n*n]=1.0;
    rep(i,1,n) rep(j,1,n)
    {
        a[t(i,j)][t(i,j)]=1.0;
        rep(x,1,n) rep(y,1,n) if(x^y&&mp[x][i]*mp[y][j])
            a[t(i,j)][t(x,y)]-=1.0*(i==x?p[x]:(1.0-p[x])/d[x])*(j==y?p[y]:(1.0-p[y])/d[y]);
    }
    Gauss(n*n);rep(i,1,n) printf("%.6lf ",a[t(i,i)][n*n]);
}