水题啦
网络流+二分
误差才10^-3，乱搞直接开longlong暴力每个都乘1000，输出时除一下就好了
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
# define int ll
const int _(210), __(1e5 + 10), INF(1e18);
IL ll Read(){
RG char c = getchar(); RG ll x = 0, z = 1;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int sum, b[_], n, m, w[__], fst[_], nxt[__], to[__], cnt, tmp1[_], tmp2[__], tcnt;
int S, T, lev[_], cur[_], max_flow;
queue <int> Q;
IL void Add(RG int u, RG int v, RG int f){
w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}
IL int Dfs(RG int u, RG int maxf){
if(u == T) return maxf;
RG int ret = 0;
for(RG int &e = cur[u]; e != -1; e = nxt[e]){
if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
RG int f = Dfs(to[e], min(w[e], maxf - ret));
ret += f; w[e ^ 1] += f; w[e] -= f;
if(ret == maxf) break;
}
return ret;
}
IL bool Bfs(){
Fill(lev, 0); lev[S] = 1; Q.push(S);
while(!Q.empty()){
RG int u = Q.front(); Q.pop();
for(RG int e = fst[u]; e != -1; e = nxt[e]){
if(lev[to[e]] || !w[e]) continue;
lev[to[e]] = lev[u] + 1;
Q.push(to[e]);
}
}
return lev[T];
}
IL bool Check(RG int x){
Copy(fst, tmp1); Copy(w, tmp2); cnt = tcnt;
for(RG int i = 1; i <= m; ++i) Add(S, i + n, x * b[i]);
for(max_flow = 0; Bfs(); ) Copy(cur, fst), max_flow += Dfs(S, INF);
return max_flow == sum;
}
# undef int
int main(RG int argc, RG char* argv[]){
# define int ll
n = Read(); m = Read(); Fill(fst, -1); T = n + m + 1;
for(RG int i = 1, a; i <= n; ++i) a = Read() * 1000, Add(i, T, a), sum += a;
for(RG int i = 1; i <= m; ++i) b[i] = Read();
for(RG int i = 1; i <= m; ++i)
for(RG int j = 1; j <= n; ++j)
if(Read()) Add(i + n, j, INF);
Copy(tmp1, fst); Copy(tmp2, w); tcnt = cnt;
RG int l = 0, r = 1e8, ans = 0;
while(l <= r){
RG int mid = (l + r) >> 1;
if(Check(mid)) ans = mid, r = mid - 1;
else l = mid + 1;
}
printf("%.3lf
", 1.0 * ans / 1000);
return 0;
}