彩票

对于⼀个已知的Ans，更新Ans的条件为：
(P1 + P2 + …… + Pk） / (T1 + T2 + …… + Tk) > Ans
变形后：P1 – T1 * Ans + P2 – T2 * Ans + …… + Pk – Tk * Ans > 0
所以将Pi – Ti * Ans作为边权，⼆分Ans，跑最长路（因为要使等式左边⼤于0）；注意存在正环的情况，⽤SPFA可判环，遇到环则return true（路径可⽆限延长所以⼀定⼤于0）；另外注意精度，保留3位⼩数，那么⼆分的条件最好为while(r - l >0.0001)

spfa：垂死病中惊坐起

（手推样例推了半天没推出来....）

 

 

#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#define ll long long
#define db double
using namespace std;
const int N = 105;
const db eps = 0.0000001;
const db inf = 999999.9;

int n,head[N],ct[N],cnt;
db dis[N],tim[N][N];
bool vis[N];
struct edge
{
    int nxt,to;
    db wei;
} e[N * N];

void add(int a,int b,db c)
{
    e[++cnt].nxt = head[a];
    e[cnt].to = b;
    e[cnt].wei = c;
    head[a] = cnt;
}

bool spfa(db k)
{
    memset(ct,0,sizeof(ct));
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= n; i++)
        dis[i] = -inf;
    queue<int> q;
    q.push(1);
    dis[1]= 0;
    while(q.size())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        ct[u]++;
        if(ct[u] == n)
            return true;
        for(int i = head[u]; i; i = e[i].nxt)
        {
            int v = e[i].to;
            db w = e[i].wei - k * tim[u][v];
            if(dis[v] < dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    return dis[n] >= 0;
}

int main()
{
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
        {
            db a;
            scanf("%lf",&a);
            if(a == 0)
                continue;
            add(i,j,a);
        }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            scanf("%lf",&tim[i][j]);
    double l = 0.0,r = 10000.0;
    while(r - l > eps)
    {
        db mid = (l + r)/2;
        if(spfa(mid))
            l = mid;
        else
            r = mid;
    }
    printf("%.3lf",l);
    return 0;

}