2014
01-26

# Destroying the bus stations

Gabiluso is one of the greatest spies in his country. Now he’s trying to complete an “impossible” mission —– to make it slow for the army of City Colugu to reach the airport. City Colugu has n bus stations and m roads. Each road connects two bus stations directly, and all roads are one way streets. In order to keep the air clean, the government bans all military vehicles. So the army must take buses to go to the airport. There may be more than one road between two bus stations. If a bus station is destroyed, all roads connecting that station will become no use. What’s Gabiluso needs to do is destroying some bus stations to make the army can’t get to the airport in k minutes. It takes exactly one minute for a bus to pass any road. All bus stations are numbered from 1 to n. The No.1 bus station is in the barrack and the No. n station is in the airport. The army always set out from the No. 1 station.
No.1 station and No. n station can’t be destroyed because of the heavy guard. Of course there is no road from No.1 station to No. n station.

Please help Gabiluso to calculate the minimum number of bus stations he must destroy to complete his mission.

There are several test cases. Input ends with three zeros.

For each test case:

The first line contains 3 integers, n, m and k. (0< n <=50, 0< m<=4000, 0 < k < 1000)
Then m lines follows. Each line contains 2 integers, s and f, indicating that there is a road from station No. s to station No. f.

There are several test cases. Input ends with three zeros.

For each test case:

The first line contains 3 integers, n, m and k. (0< n <=50, 0< m<=4000, 0 < k < 1000)
Then m lines follows. Each line contains 2 integers, s and f, indicating that there is a road from station No. s to station No. f.

5 7 3
1 3
3 4
4 5
1 2
2 5
1 4
4 5
0 0 0

2

#include<iostream>
#include<queue>
using namespace std;
struct Node{
int flow,cost;
int next;
};
Node edges[1000000];
int loc;
int s,t;
}
int n,r,k;
bool visit[100];
int dis[100];
int father[100];
bool spfa(){
memset(dis,-1,sizeof(dis));
memset(father,-1,sizeof(father));
memset(visit,false,sizeof(visit));
int now=s;
dis[now]=0;
visit[now]=true;
queue<int> qu;
qu.push(now);
while(!qu.empty()){
now=qu.front(),qu.pop();
visit[now]=false;
int p;
if(edges[p].flow>0){
int temp=dis[now]+edges[p].cost;
}
}
}
}
}
return dis[t]>0&&dis[t]<=k;
}
int mincost(){
int flow=0;
while(spfa()){
int end=father[t];
int min=1000000;
while(end!=-1){
min=edges[end].flow<min?edges[end].flow:min;
end=father[edges[end].here];
}
flow+=min;
end=father[t];
while(end!=-1){
edges[end].flow-=min;
edges[end^1].flow+=min;
end=father[edges[end].here];
}
}
return flow;
}
int main(){
while(scanf(“%d%d%d”,&n,&r,&k),n){
s=1,t=2*n-2;
loc=0;
for(int i=1;i<n-1;i++){
}
for( i=1;i<=r;i++){
int a,b;
scanf(“%d%d”,&a,&b);
a–,b–;