BZOJ 3143Hnoi2013游走 期望+高斯消元

Posted 2020-06-14

如果纯模拟，就会死循环，而随着循环每个点的期望会逼近一个值，高斯消元就通过列方正组求出这个值。

#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
const double eps=1e-9;
bool vis[503];
double f[503],a[503][503],ans[500*500];
int N,M,cnt=0,du[503],a1[500*500],a2[500*500];
double fabs(double x){return x>0?x:-x;}
int getint(){char c;while (!isdigit(c=getchar()));int a=c-‘0‘;while(isdigit(c=getchar()))a=a*10+c-‘0‘;return a;}
void prepare(){
    for(int i=1;i<=M;++i){
        a[a1[i]][a2[i]]+=1.0/du[a2[i]];
        a[a2[i]][a1[i]]+=1.0/du[a1[i]];
    }
    for(int i=1;i<=N;++i)a[N][i]=0;
    for(int i=1;i<N;++i)a[i][i]=-1.0;
    a[1][N+1]=-1.0;a[N][N]=1.0;
}
void swapp(double &x,double &y){double z=x;x=y;y=z;}
void gauss(){
    for(int i=1;i<=N;++i){
        int now=i;
        for(int j=i+1;j<=N;++j)if(fabs(a[j][i])>fabs(a[now][i]))now=j;
        if (now!=i)for(int j=i;j<=N+1;++j)swapp(a[now][j],a[i][j]);
        for(int j=i+1;j<=N;++j){
            double t=a[j][i]/a[i][i];
            for(int k=i;k<=N+1;++k)a[j][k]-=t*a[i][k];
        }
    }
    for(int i=N;i>=1;--i){
        for(int j=N;j>i;--j){
            a[i][N+1]-=a[j][N+1]*a[i][j];
        }a[i][N+1]/=a[i][i];
    }
}
bool cmp(double a,double b){return a>b;}
int main(){
    memset(a,0,sizeof(a));
    memset(du,0,sizeof(du));
    N=getint();M=getint();
    for(int i=1;i<=M;++i){
        a1[i]=getint();a2[i]=getint();
        du[a1[i]]++;du[a2[i]]++;
    }prepare();
    gauss();
    cnt=0;
    for(int i=1;i<=M;++i){
        ans[++cnt]=a[a1[i]][N+1]/du[a1[i]]+a[a2[i]][N+1]/du[a2[i]];
    }
    sort(ans+1,ans+M+1,cmp);
    double sa=0;
    for(int i=1;i<=M;++i) sa+=ans[i]*i*1.0;
    printf("%.3lf\n",sa);
    return 0;
}

　　这样就可以了