然而这是一道网络流。。。
如果满足Bob，使总费用最大：
设最大流的每条边流量(不是容量)为w[i]，分配到每条边的费用为p[i]，最大流量为wmax，p[i]的和为P
那么显然w[i] * p[i]的和小于等于wmax * P
证明：
[wmax * P = sum wmax * p[i].....................(1)
]
[(1) - sum w[i]*p[i] = sum (wmax - w[i]) * p[i] ge 0
]
证毕
那么如果满足Alice，使总费用最小
就只要使得最大流中最大的流量的边的流量最小
于是二分这个最小流量，把所有边的容量对它取min后跑一遍容量为分数的最大流，与原本的最大流比较即可
# 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;
const int _(110), __(2010), INF(2147483647);
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 n, m, fst[_], nxt[__], to[__], cnt, A[__], B[__], p, S, T, lev[_], cur[_];
double C[__], w[__], max_flow, ans;
queue <int> Q;
IL void Add(RG int u, RG int v, RG double 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 double Dfs(RG int u, RG double maxf){
if(u == T) return maxf;
RG double ret = 0;
for(RG int &e = cur[u]; e != -1; e = nxt[e]){
if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
RG double f = Dfs(to[e], min(w[e], maxf - ret));
ret += f; w[e ^ 1] += f; w[e] -= f;
if(ret == maxf) break;
}
if(!ret) lev[u] = 0;
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 double Check(RG double x){
Fill(fst, -1); cnt = 0;
for(RG int i = 1; i <= m; i++) Add(A[i], B[i], min(C[i], x));
for(max_flow = 0; Bfs(); ) Copy(cur, fst), max_flow += Dfs(S, INF);
return max_flow;
}
int main(RG int argc, RG char* argv[]){
n = Read(); m = Read(); p = Read(); S = 1; T = n;
for(RG int i = 1; i <= m; i++) A[i] = Read(), B[i] = Read(), C[i] = Read();
ans = Check(INF);
RG double l = 0, r = 1000000;
while(r - l >= 1e-6){
RG double mid = (l + r) / 2;
if(ans == Check(mid)) r = mid;
else l = mid;
}
printf("%.0lf
%.4lf
", ans, l * p);
return 0;
}