[SDOI2013]费用流

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\n%.4lf\n", ans, l * p);
    return 0;
}