對於一張給定的運輸網絡,Alice 先確定一個最大流,如果有多種解,Alice 可以任選一種;之後 Bob 在每條邊上分配單位花費(單位花費必須是非負實數),要求所有邊的單位花費之和等於
需要注意到,Bob 在分配單位花費之前,已經知道 Alice 所給出的最大流方案。
現茌 Alice 希望總費用盡量小,而 Bob 希望總費用盡量大。我們想知道,如果兩個人都執行最優策略,最大流的值和總費用分別爲多少。
爲了簡化問題,我們假設源點
題解:顯然 Bob 會把所有費用都放在 Alice 流量最大的一條邊上,那麼我們只要知道 Alice 最大流流量最大的邊最小是多少,我們二分一下上界,就變成判定性問題了。然而一直卡在
#include<bits/stdc++.h>
const int N = 105;
const int M = 1050;
const double INF = 1e8;
const double eps = 1e-5;
template <typename T> void read(T &x) {
x = 0; char c = getchar();
for (; !isdigit(c); c = getchar());
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
}
int n, m, con, S, T, a[M], b[M], first[N], s, q[N], h[N], cur[N], MAX_FLOW, shit;
double c[M];
struct edge {
int y, next;
double v;
}mp[M*2];
void ins(int x, int y, double v) {
//printf("ins x=%d y=%d v=%d\n", x, y, v);
mp[++s] = (edge) {y, first[x], v}; first[x] = s;
mp[++s] = (edge) {x, first[y], 0}; first[y] = s;
}
void build(double UP) {
memset(first, 0, sizeof(first)); s = 1;
for (int i=1; i <= m; i++)
ins(a[i], b[i], std::min(c[i], UP));
}
bool bfs() {
for (int i=1; i <= N; i++)
h[i] = 0, cur[i] = first[i];
int head = 1, tail = 1;
h[q[head] = S] = 1;
for (int x=q[head]; head <= tail; x=q[++head])
for (int t=first[x]; t; t=mp[t].next)
if (mp[t].v > eps && !h[mp[t].y]){
h[mp[t].y] = h[x] + 1,
q[++tail] = mp[t].y;
if (mp[t].y == T) return 1;
}
return 0;
}
double dfs(int x, double f) {
if (x == T) return f;
double used = 0, b;
for (int t=cur[x]; t; t=cur[x]=mp[t].next)
if (h[x] + 1 == h[mp[t].y]) {
b = dfs(mp[t].y, std::min(mp[t].v, f - used));
mp[t].v -= b;
mp[t^1].v += b;
used += b;
if (fabs(used - f) < eps) return used;
}
h[x] = -1;
return used;
}
double Dinic(double UP) {
build(UP);
double ret = 0;
while (bfs()) ret += dfs(S, INF);
return ret;
}
int main() {
scanf("%d%d%d", &n, &m, &con);
for (int i=1; i <= m; i++)
read(a[i]), read(b[i]), scanf("%lf", &c[i]);
S = 1, T = n;
printf("%d\n", MAX_FLOW = (int)Dinic(INF));
double l = 0, r = INF;
while (l + eps < r) {
double mid = (l + r) / 2;
if (fabs(Dinic(mid) - MAX_FLOW) < eps)
r = mid;
else
l = mid;
}
printf("%.3f\n", l * con);
return 0;
}