[USACO07DEC]Sightseeing Cows
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。
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<cstdlib> 7 #include<cctype> 8 #include<stack> 9 #include<queue> 10 #include<vector> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-4; 20 const int maxn = 1e3 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), las = ' '; 25 while(!isdigit(ch)) las = ch, ch = getchar(); 26 while(isdigit(ch)) ans = (ans << 3) + (ans << 1) + ch - '0', ch = getchar(); 27 if(las == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) putchar('-'), x = -x; 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m, val[maxn]; 38 vector<int> v[maxn], c[maxn]; 39 40 bool in[maxn]; 41 db dis[maxn]; 42 int cnt[maxn]; 43 bool judge(db x) 44 { 45 Mem(in, 0); Mem(cnt, 0); Mem(dis, 0); 46 queue<int> q; 47 for(int i = 1; i <= n; ++i) q.push(i); 48 while(!q.empty()) 49 { 50 int now = q.front(); q.pop(); in[now] = 0; 51 for(int i = 0; i < (int)v[now].size(); ++i) 52 { 53 if(dis[v[now][i]] > x * c[now][i] - val[v[now][i]] + dis[now]) 54 { 55 dis[v[now][i]] = x * c[now][i] - val[v[now][i]] + dis[now]; 56 if(!in[v[now][i]]) 57 { 58 if(++cnt[v[now][i]] == n - 1) return 1; 59 q.push(v[now][i]); 60 } 61 } 62 } 63 } 64 return 0; 65 } 66 67 int main() 68 { 69 n = read(); m = read(); 70 for(int i = 1; i <= n; ++i) val[i] = read(); 71 for(int i = 1; i <= m; ++i) 72 { 73 int x = read(), y = read(), co = read(); 74 v[x].push_back(y); c[x].push_back(co); 75 } 76 db L = 0, R = 1e6; 77 while(R - L > eps) 78 { 79 db mid = (L + R) / 2; 80 if(judge(mid)) L = mid; 81 else R = mid; 82 } 83 if(L < eps) write(0), enter; 84 else printf("%.2lf\n", L); 85 return 0; 86 }
