Secret Milking Machine
Description
Farmer
John is constructing a new milking machine and wishes to keep it secret
as long as possible. He has hidden in it deep within his farm and needs
to be able to get to the machine without being detected. He must make a
total of T (1 <= T <= 200) trips to the machine during its
construction. He has a secret tunnel that he uses only for the return
trips.
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
* Line 1: Three space-separated integers: N, P, and T
* 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
* Line 1: A single integer that is the minimum possible length of the longest segment of Farmer John's route.
Sample Input
7 9 2
1 2 2
2 3 5
3 7 5
1 4 1
4 3 1
4 5 7
5 7 1
1 6 3
6 7 3
Sample Output
5
Hint
Farmer
John can travel trails 1 - 2 - 3 - 7 and 1 - 6 - 7. None of the trails
travelled exceeds 5 units in length. It is impossible for Farmer John
to travel from 1 to 7 twice without using at least one trail of length
5.
Huge input data,scanf is recommended.
題意：fj想從一個點a到另外一個點b，在每條邊只走一次的情況下，走k次從a到b，使得經過的邊最大的盡量小。
分析：又是二分，這個題就是圖的邊連通度，其每條通過流量的路徑被稱為弱獨立軌。原圖中u->v，如果是有向圖，則從u到v一條容量為1的邊。如果是無向圖，則從u到v一條容量為1的邊，從v到u一條容量為1的邊。這樣保證每條邊只經過一個流量，而每個點可以被通關過多次，只要這個點有多個入度跟出度。就這樣了，最大流。
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#define Min(a, b) (a) < (b) ? a : b
#define Max(a, b) (a) > (b) ? a : b
using namespace std;
const int MAXN = 1005;
const int MAXM = 210000;
const int INF = 1100000000;
struct Edge
{
int st, ed;
int next;
int flow;
int cap;
}edge[MAXM];
int head[MAXN], level[MAXN], que[MAXN], E;
struct List
{
int u, v, w;
}list[40000];
void add(int u, int v, int w)
{
//printf("add %d %d %d\n", u, v, w);
edge[E].flow = 0;
edge[E].cap = w;
edge[E].st = u;
edge[E].ed = v;
edge[E].next = head[u];
head[u] = E++;
edge[E].flow = 0;
edge[E].cap = 0;
edge[E].st = v;
edge[E].ed = u;
edge[E].next = head[v];
head[v] = E++;
}
int dinic_bfs(int src, int dest, int ver)
{
int i, j;
for (i = 0; i <= ver; i++)
{
level[i] = -1;
}
int rear = 1;
que[0] = src; level[src] = 0;
for(i = 0; i < rear; i++)
{
for(j = head[que[i]]; j != -1; j = edge[j].next)
{
if(level[edge[j].ed] == -1 && edge[j].cap > edge[j].flow)
{
level[edge[j].ed] = level[que[i]]+1;
que[rear++] = edge[j].ed;
}
}
}
return level[dest] >= 0;
}
int dinic_dfs(int src, int dest, int ver)
{
int stk[MAXN], top = 0;
int ret = 0, cur, ptr, pre[MAXN], minf, i;
int del[MAXN];
for (i = 0; i <= ver; i++)
{
del[i] = 0;
}
stk[top++] = src;
pre[src] = src;
cur = src;
while(top)
{
while(cur != dest && top)
{
for(i = head[cur]; i != -1; i = edge[i].next)
{
if(level[edge[i].ed] == level[cur] + 1 && edge[i].cap > edge[i].flow && !del[edge[i].ed])
{
stk[top++] = edge[i].ed;
cur = edge[i].ed;
pre[edge[i].ed] = i;
break;
}
}
if(i == -1)
{
del[cur] = 1;
top--;
if(top) cur = stk[top-1];
}
}
if(cur == dest)
{
minf = INF;
while(cur != src)
{
cur = pre[cur];
if(edge[cur].cap - edge[cur].flow < minf) minf = edge[cur].cap - edge[cur].flow;
cur = edge[cur].st;
}
cur = dest;
while(cur != src)
{
cur = pre[cur];
edge[cur].flow += minf;
edge[cur^1].flow -= minf;
if(edge[cur].cap - edge[cur].flow == 0)
{
ptr = edge[cur].st;
}
cur = edge[cur].st;
}
while(top > 0&& stk[top-1] != ptr) top--;
if(top) cur = stk[top-1];
ret += minf;
}
}
return ret;
}
int Dinic(int src, int dest, int ver)
{
int ret = 0, t;
while(dinic_bfs(src, dest, ver))
{
t = dinic_dfs(src, dest, ver);
if(t) ret += t;
else break;
}
return ret;
}
void build (int length, int n, int m)
{
E = 0;
int i, u, v, w;
int s = 0, t = n + 1, ver = t + 1;
for (i = 0; i <= ver; i++)
{
head[i] = -1;
}
for (i = 0; i < m; i++)
{
u = list[i].u, v = list[i].v, w = list[i].w;
if (w <= length)
{
add(u, v, 1);
add(v, u, 1);
}
}
add(s, 1, INF);
add(n, t, INF);
}
int main()
{
int n, m, k, i, max, s, t, ver, l, r, mid, flow;
scanf("%d%d%d", &n, &m, &k);
for (i = 0, max = 0; i < m; i++)
{
scanf("%d%d%d", &list[i].u, &list[i].v, &list[i].w);
max = Max(max, list[i].w);
}
s = 0, t = n + 1, ver = t + 1;
l = 0, r = max + 1;
while (l < r)
{
mid = (l + r) >> 1;
build(mid, n, m);
flow = Dinic(s, t, ver);
if (flow >= k)
{
r = mid;
}
else
{
l = mid + 1;
}
}
printf("%d\n", r);
return 0;
}
/*
*/