BZOJ3993 [SDOI2015]星际战争

二分答案。。。然后最大流验证是否可行。。。

没了，好水啊QAQ

 /**************************************************************
     Problem: 3993
     User: rausen
     Language: C++
     Result: Accepted
     Time:40 ms
     Memory:1156 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 typedef double lf;
 const int N = ;
 const lf inf = 1e10;
 const lf eps = 1e-;
 const int M = N * N << ;

 inline int read();

 struct edge {
     int next, to;
     lf f;
     edge(int _n = , int _t = , lf _f = 0.0) : next(_n), to(_t), f(_f) {}
 } e[M];

 int n, m, S, T;
 int b[N];
 int first[N], tot = ;
 int q[N], d[N];

 inline void Add_Edges(int x, int y, lf z) {
     e[++tot] = edge(first[x], y, z), first[x] = tot;
     e[++tot] = edge(first[y], x, ), first[y] = tot;
 }

 #define y e[x].to
 #define p q[l]
 bool bfs() {
     static int l, r, x;
     memset(d, -, sizeof(d));
     d[q[] = S] = ;
     for (l = r = ; l != r + ; ++l)
         for (x = first[p]; x; x = e[x].next)
             if (!~d[y] && e[x].f > eps) {
                 d[q[++r] = y] = d[p] + ;
                 if (y == T) return ;
             }
     return ;
 }
 #undef p

 lf dfs(int p, lf lim) {
   if (p == T || !lim) return lim;
   int x;
   lf tmp, rest = lim;
   for (x = first[p]; x && rest > eps; x = e[x].next)
     if (d[y] == d[p] +  && ((tmp = min(e[x].f, rest)) > eps)) {
       rest -= (tmp = dfs(y, tmp));
       e[x].f -= tmp, e[x ^ ].f += tmp;
       if (!rest) return lim;
     }
   if (rest > eps) d[p] = -;
   return lim - rest;
 }

 inline bool check(lf t) {
     static int x;
     for (x = ; x <= tot; x += )
         e[x].f += e[x ^ ].f, e[x ^ ].f = ;
     for (x = first[S]; x; x = e[x].next)
         e[x].f = b[y - n] * t;
     while (bfs()) dfs(S, inf);
     for (x = first[T]; x; x = e[x].next)
         if (e[x ^ ].f > eps) return ;
     return ;
 }
 #undef y

 int main() {
     int i, j;
     lf l, r, mid;
     n = read(), m = read();
     S = n + m + , T = S + ;
     for (i = ; i <= n; ++i)
         Add_Edges(i, T, read());
     for (i = ; i <= m; ++i)
         b[i] = read(), Add_Edges(S, n + i, );
     for (i = ; i <= m; ++i)
         for (j = ; j <= n; ++j)
             if (read() == ) Add_Edges(n + i, j, inf);
     l = , r = 5e6;
     while (r - l > eps) {
         mid = (l + r) / 2.0;
         if (check(mid)) r = mid;
         else l = mid;
     }
     printf("%.6lf\n", (l + r) / 2.0);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }