Description
The farm comprises N (2 <= N <= 200) landmarks (numbered 1..N) connected by P (1 <= P <= 40,000) bidirectional trails (numbered 1..P) and with a positive length that does not exceed 1,000,000. Multiple trails might join a pair of landmarks.
To minimize his chances of detection, FJ knows he cannot use any trail on the farm more than once and that he should try to use the shortest trails.
Help FJ get from the barn (landmark 1) to the secret milking machine (landmark N) a total of T times. Find the minimum possible length of the longest single trail that he will have to use, subject to the constraint that he use no trail more than once. (Note well: The goal is to minimize the length of the longest trail, not the sum of the trail lengths.)
It is guaranteed that FJ can make all T trips without reusing a trail.
Input
* Lines 2..P+1: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, indicating that a trail connects landmark A_i to landmark B_i with length L_i.
Output
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std; const int MAXN = 210;
const int MAXE = 40010 * 2;
const int INF = 0x7f7f7f7f; struct Dinic {
int n, m, st, ed, ecnt, maxlen;
int head[MAXN];
int cur[MAXN], d[MAXN];
int to[MAXE], next[MAXE], flow[MAXE], cap[MAXE], len[MAXE]; void init(int ss, int tt) {
st = ss; ed = tt;
ecnt = 2;
memset(head, 0, sizeof(head));
} void add_edge(int u, int v, int c, int l) {
len[ecnt] = l; to[ecnt] = v; cap[ecnt] = c; flow[ecnt] = 0; next[ecnt] = head[u]; head[u] = ecnt++;
len[ecnt] = l; to[ecnt] = u; cap[ecnt] = c; flow[ecnt] = 0; next[ecnt] = head[v]; head[v] = ecnt++;
} bool bfs() {
memset(d, 0, sizeof(d));
queue<int> que; que.push(st);
d[st] = 1;
while(!que.empty()) {
int u = que.front(); que.pop();
for(int p = head[u]; p; p = next[p]) {
if(len[p] > maxlen) continue;
int v = to[p];
if(!d[v] && cap[p] > flow[p]) {
d[v] = d[u] + 1;
que.push(v);
if(v == ed) return true;
}
}
}
return d[ed];
} int dfs(int u, int a) {
if(u == ed || a == 0) return a;
int outflow = 0, f;
for(int &p = cur[u]; p; p = next[p]) {
if(len[p] > maxlen) continue;
int v = to[p];
if(d[u] + 1 == d[v] && (f = dfs(v, min(a, cap[p] - flow[p]))) > 0) {
flow[p] += f;
flow[p ^ 1] -= f;
outflow += f;
a -= f;
if(a == 0) break;
}
}
return outflow;
} int Maxflow(int mlen) {
int ans = 0; maxlen = mlen;
while(bfs()) {
for(int i = 0; i <= ed; ++i) cur[i] = head[i];
ans += dfs(st, INF);
}
return ans;
}
} G; int main() {
int n, m, T, left = 0, right = 0;
scanf("%d%d%d", &n, &m, &T);
G.init(1, n);
for(int i = 0; i < m; ++i) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
G.add_edge(a, b, 1, c);
if(right < c) right = c;
}
while(left < right) {
memset(G.flow, 0, sizeof(G.flow));
int mid = (left + right) >> 1;
if(G.Maxflow(mid) < T) left = mid + 1;
else right = mid;
}
printf("%d\n", right);
}