[USACO07DEC]Sightseeing Cows

嘟嘟嘟

这题好像属于01分数规划问题，叫什么最优比率生成环。

题目概括一下，就是求一个环，满足∑v[i] / ∑c[i]最大。

我们可以堆上面的式子变个型：令 x = ∑v[i] / ∑c[i],则x * ∑c[i] = v[i] => ∑x * c[i] - v[i] = 0。于是对于任何能取到的x',满足∑x * c[i] - v[i] <= 0；对于不能取到的x', ∑x * c[i] - v[i] > 0。

于可以实数二分答案，用spfa判断负环。

然后实数二分又RE了……调了好就还是看了题解。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #include<cctype>
 #include<stack>
 #include<queue>
 #include<vector>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e3 + ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), las = ' ';
   while(!isdigit(ch)) las = ch, ch = getchar();
   while(isdigit(ch)) ans = (ans << ) + (ans << ) + ch - '', ch = getchar();
   if(las == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) putchar('-'), x = -x;
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, m, val[maxn];
 vector<int> v[maxn], c[maxn];

 bool in[maxn];
 db dis[maxn];
 int cnt[maxn];
 bool judge(db x)
 {
   Mem(in, ); Mem(cnt, ); Mem(dis, );
   queue<int> q;
   for(int i = ; i <= n; ++i) q.push(i);
   while(!q.empty())
     {
       int now = q.front(); q.pop(); in[now] = ;
       for(int i = ; i < (int)v[now].size(); ++i)
     {
       if(dis[v[now][i]] > x * c[now][i] - val[v[now][i]] + dis[now])
         {
           dis[v[now][i]] = x * c[now][i] - val[v[now][i]] + dis[now];
           if(!in[v[now][i]])
         {
           if(++cnt[v[now][i]] == n - ) return ;
           q.push(v[now][i]);
         }
         }
     }
     }
   return ;
 }

 int main()
 {
   n = read(); m = read();
   for(int i = ; i <= n; ++i) val[i] = read();
   for(int i = ; i <= m; ++i)
     {
       int x = read(), y = read(), co = read();
       v[x].push_back(y); c[x].push_back(co);
     }
   db L = , R = 1e6;
   while(R - L > eps)
     {
       db mid = (L + R) / ;
       if(judge(mid)) L = mid;
       else R = mid;
     }
   if(L < eps) write(), enter;
   else printf("%.2lf\n", L);
   return ;
 }