mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
125 lines
3.0 KiB
C++
125 lines
3.0 KiB
C++
#include <stdio.h>
|
|
#include <string.h>
|
|
#include <iostream>
|
|
#include <algorithm>
|
|
#include <queue>
|
|
#include <map>
|
|
#include <set>
|
|
#include <vector>
|
|
#include <string>
|
|
#include <math.h>
|
|
using namespace std;
|
|
#define eps 1e-6
|
|
const int MAXN=220;
|
|
double a[MAXN][MAXN],x[MAXN];
|
|
int equ,var;
|
|
int Gauss()
|
|
{
|
|
int i,j,k,col,max_r;
|
|
for(k=0,col=0;k<equ&&col<var;k++,col++){
|
|
max_r = k;
|
|
for(i=k+1;i<equ;i++)
|
|
if(fabs(a[i][col])>fabs(a[max_r][col]))
|
|
max_r = i;
|
|
if(fabs(a[max_r][col])<eps)return 0;
|
|
if(k != max_r){
|
|
for(j=col;j<var;j++)
|
|
swap(a[k][j],a[max_r][j]);
|
|
swap(x[k],x[max_r]);
|
|
}
|
|
x[k]/=a[k][col];
|
|
for(j=col+1;j<var;j++)a[k][j]/=a[k][col];
|
|
a[k][col] = 1;
|
|
for(i=0;i<equ;i++)
|
|
if(i!=k){
|
|
x[i] -= x[k]*a[i][k];
|
|
for(j=col+1;j<var;j++)a[i][j]-=a[k][j]*a[i][col];
|
|
a[i][col]=0;
|
|
}
|
|
}
|
|
return 1;
|
|
}
|
|
vector<int>vec[MAXN];
|
|
int g[MAXN][MAXN];
|
|
int du[MAXN];
|
|
int add[MAXN];
|
|
int main()
|
|
{
|
|
int T;
|
|
int n,m;
|
|
scanf("%d",&T);
|
|
while(T--)
|
|
{
|
|
scanf("%d%d",&n,&m);
|
|
for(int i = 0;i < n;i++)
|
|
vec[i].clear();
|
|
memset(g,0,sizeof(g));
|
|
memset(du,0,sizeof(du));
|
|
int u,v;
|
|
while(m--)
|
|
{
|
|
scanf("%d%d",&u,&v);
|
|
if(u == v)continue;
|
|
g[u][v] = 1;
|
|
}
|
|
for(int i = 0;i < n;i++)
|
|
{
|
|
for(int j = 0;j < n;j++)
|
|
if(j != i && g[i][j])
|
|
{
|
|
du[i]++;
|
|
vec[j].push_back(i);
|
|
}
|
|
}
|
|
equ = var = n;
|
|
for(int i = 0;i < n;i++)
|
|
x[i] = 0;
|
|
memset(a,0,sizeof(a));
|
|
for(int i = 0;i < n;i++)
|
|
{
|
|
a[i][i] = -1;
|
|
int sz = vec[i].size();
|
|
for(int j = 0;j < sz;j++)
|
|
{
|
|
int v = vec[i][j];
|
|
if(i == v)continue;
|
|
a[i][v] = 1.0 / du[v];
|
|
}
|
|
}
|
|
for(int i = 0;i < n;i++)
|
|
a[n-1][i] = 1;
|
|
x[n-1] = 1;
|
|
for(int k = 0;k < n-1;k++)
|
|
if(g[n-1][k] == 0)
|
|
{
|
|
for(int i = 0;i < n-1;i++)
|
|
{
|
|
if(g[n-1][i])a[i][var] = 1.0/(du[n-1]+1);
|
|
else a[i][var] = 0;
|
|
}
|
|
a[k][var] = 1.0/(du[n-1]+1);
|
|
a[n-1][var] = 1;
|
|
add[var] = k;
|
|
var++;
|
|
}
|
|
if(!Gauss())
|
|
{
|
|
printf("INF\n");
|
|
continue;
|
|
}
|
|
double tt = x[n-1];
|
|
double now = x[n-1];
|
|
int ans = -1;
|
|
for(int i = n;i < var;i++)
|
|
{
|
|
if(x[n-1]/a[n-1][i] > now)
|
|
{
|
|
ans = add[i];
|
|
now = x[n-1]/a[n-1][i];
|
|
}
|
|
}
|
|
printf("%d %d\n",1,ans);
|
|
}
|
|
return 0;
|
|
}
|