POJ2455 Secret Milking Machine
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12324 | Accepted: 3589 |
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
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
Huge input data,scanf is recommended.
Source
————————————————————————————————————
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset> using namespace std; #define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 500 struct node
{
int u, v, next, cap;
} edge[MAXN*MAXN],mp[MAXN*MAXN];
int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN],p[MAXN];
int cnt;
int ct;
int n,m,k; int N;
void init()
{
cnt = 0;
memset(s, -1, sizeof(s));
} void add(int u, int v, int c)
{
edge[cnt].u = u;
edge[cnt].v = v;
edge[cnt].cap = c;
edge[cnt].next = s[u];
s[u] = cnt++;
edge[cnt].u = v;
edge[cnt].v = u;
edge[cnt].cap = c;
edge[cnt].next = s[v];
s[v] = cnt++;
} bool BFS(int ss, int ee)
{
memset(d, 0, sizeof d);
d[ss] = 1;
queue<int>q;
q.push(ss);
while (!q.empty())
{
int pre = q.front();
q.pop();
for (int i = s[pre]; ~i; i = edge[i].next)
{
int v = edge[i].v;
if (edge[i].cap > 0 && !d[v])
{
d[v] = d[pre] + 1;
q.push(v);
}
}
}
return d[ee];
} int DFS(int x, int exp, int ee)
{
if (x == ee||!exp) return exp;
int temp,flow=0;
for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
{
int v = edge[i].v;
if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
{
edge[i].cap -= temp;
edge[i ^ 1].cap += temp;
flow += temp;
exp -= temp;
if (!exp) break;
}
}
if (!flow) d[x] = 0;
return flow;
} int Dinic_flow(int mid)
{
init();
for(int i=1; i<=n; i++)
for(int j=p[i]; ~j; j=mp[j].next)
if(mp[j].cap<=mid)
add(mp[j].u,mp[j].v,1);
int ss=1,ee=n;
int ans = 0;
while (BFS(ss, ee))
{
for (int i = 0; i <=ee; i++) nt[i] = s[i];
ans+= DFS(ss, INF, ee);
}
return ans;
} int main()
{
int u,v,c;
while(~scanf("%d%d%d",&n,&m,&k))
{
ct=0;
memset(p,-1,sizeof p);
for(int i=0; i<m; i++)
{
scanf("%d%d%d",&u,&v,&c);
mp[ct].u=u;
mp[ct].v=v;
mp[ct].cap=c;
mp[ct].next=p[u];
p[u]=ct++;
}
int l=0,r=INF;
int ans=0;
while(l<=r)
{
int mid=(l+r)>>1;
if(Dinic_flow(mid)>=k) ans=mid,r=mid-1;
else l=mid+1;
}
printf("%d\n",ans);
}
return 0;
}
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