uva10828

https://vjudge.net/problem/UVA-10828

裸高斯消元。。。

但是要判无解和无穷解。

当出现一个环时会无解，环上每个点只有一个出边。

#include
using namespace std;
const int N = 110;
const double eps = 1e-8;
int n, q;
double a[N][N], d[N];
vector<int> G[N];
int mark[N];
void build()
{
    a[1][n + 1] = -1.0;
    for(int i = 1; i <= n; ++i)
    {
        a[i][i] = -1.0;
        for(int j = 0; j < G[i].size(); ++j) 
            a[i][G[i][j]] += 1.0 / d[G[i][j]];
    }
}
void gauss_jordan()
{
    for(int now = 1; now <= n; ++now)
    {
        int x = now;
        for(int i = now + 1; i <= n; ++i) if(fabs(a[i][now]) > fabs(a[x][now])) x = i;
        if(fabs(a[x][now]) < eps) continue; 
        for(int i = 1; i <= n + 1; ++i) swap(a[now][i], a[x][i]);
        double t = a[now][now];        
        for(int i = now; i <= n + 1; ++i) a[now][i] /= t;
        for(int i = 1; i <= n; ++i) if(i != now)
        {
            double t = a[i][now];
            for(int j = now; j <= n + 1; ++j) a[i][j] -= a[now][j] * t;
        }
    }
}
int main()
{
    int T = 0;
    while(scanf("%d", &n))
    {
        ++T;
        if(n == 0) break;
        memset(d, 0, sizeof(d));
        memset(a, 0, sizeof(a));
        memset(mark, 0, sizeof(mark));        
        while(1)
        {
            int a, b; scanf("%d%d", &a, &b);
            if(a == 0 && b == 0) break;
            G[b].push_back(a);
            d[a] += 1.0;
        }
        build();
        gauss_jordan();
        for(int i = n; i; --i) 
        {
            if(fabs(a[i][i]) < eps && fabs(a[i][n + 1]) > eps)
                mark[i] = 1;            
            for(int j = i + 1; j <= n; ++j) if(fabs(a[i][j]) > eps && mark[j]) mark[i] = 1;
        }
        scanf("%d", &q);
        printf("Case #%d:\n", T);
        while(q--)
        {
            int x; scanf("%d", &x);
            if(mark[x]) puts("infinity");
            else if(fabs(a[x][x]) < eps) puts("0.000");
            else printf("%.3f\n", fabs(a[x][n + 1] / a[x][x]));
        }
        for(int i = 1; i <= n; ++i) G[i].clear();
    }
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/19992147orz/p/6815182.html