E |
Frequency Hopping |
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Input: Standard Input Output: Standard Output |
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20th July, 1942 Colonel Al Pacheno, According to the previous order “ref: 232/UK44i/334sda#nh$X3y”, you are required back in the DOI (Department of intelligence, London) to head the special programming contingent immediately. You are to assign a programmer for the job whose specification is attached with this letter. Level 3 Secrecy must be maintained.
Sincerely, General Shaan Konary Director, DOI London Ps: Sorry to ruin your Caribbean holiday
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232/UK44i/334sda#nh$X3y/Appx-301a
At this moment, through out Europe, our base station numbers 1 to N are actively operational through wireless channels. Immediately we require sending C secret message fragments from our head quarters (base station 1) to Nth base station. Germans have developed Zämmhäim – a machine which jams the frequency channel between base stations after a station has sent a message fragment. In that case, the base stations must transmit using a different frequency channel for each message fragment. There are several unidirectional channels set up between base stations at this moment. We can only make arrangements to set up number of frequency channels only between two base stations. Your task is to check whether all the message fragments can be sent to the desired base station with or without increasing frequency channel between any two particular base stations. You have to give us all possible options if it is required to increase frequency channel between two stations.
--End of Attachment
As members of Secret Programmers Group (SPG) you are assigned to solve this problem within 5 hrs and deliver the solution directly to Colonel Al Pacheno. You have to maintain Level 3 secrecy and destroy all documents corresponding to this as soon as you deliver the solution.
Input:
There will be multiple test cases. The first line of each test case contains three numbers N, E and C where N (0<N<101) represents the number of base stations, E (E<10000) represents the number of available connections between the base stations and C (C<2000000000) represents the number of secret message fragments that are required to send from station 1 to station N. After that, there will be E lines. Each line contains 3 numbers: b1(0<b1<101), b2(0<b2<101) and fp (0<fp<5001) which represent the number of frequency channels available currently from b1 to b2. Input is terminated when N=E=C=0.
Output:
For each test case, there will be one line of output. First, you have to print the case number. If it is possible to send C secret message fragments from the current status the output will be “possible”. Otherwise, you have to print all pairs of stations (in ascending order) if it is possible send the required message fragments by increasing the frequency channel between any one of them. If it is still impossible, you have to print “not possible”.
Sample Input Output for Sample Input
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4 4 5 1 2 5 1 3 5 2 4 5 3 4 5 4 4 5 1 2 1 1 3 5 2 4 5 3 4 1 4 4 5 1 2 1 1 3 1 2 4 1 3 4 1 0 0 0
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Case 1: possible Case 2: possible option:(1,2),(3,4) Case 3: not possible
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Problemsetter: Syed Monowar Hossain
Special Thanks: Abdullah al Mahmud
最大流， 求最小割，然后枚举最小割中弧，改其容量为C，看是否能使得最大流达到C，并输出可能的方案，抄白书:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))
#define INF 2000000000
#define MAXN 111
#define MAXM 22222
struct edge{
int u, v, cap, flow, nxt;
}e[MAXM];
int h[MAXN], cc;
void add(int u, int v, int cap, int flow){
e[cc]=(edge){u, v, cap, flow, h[u]};
h[u]=cc++;
}
void reduce(){ for(int i=0; i<cc; i++) e[i].cap-=e[i].flow; }
bool vis[MAXN];
bool operator <(const edge &x, const edge &y){
return x.u<y.u || (x.u==y.u&&x.v<y.v);
}
int d[MAXN], s, t, cur[MAXN];
void bfs(){
queue<int> q;
// cout<<t<<endl;
memset(vis, 0, sizeof(vis)); vis[t]=true;
d[t]=0; q.push(t);
while(!q.empty()){
int u=q.front(); q.pop();
// cout<<u<<endl;
for(int i=h[u]; i!=-1; i=e[i].nxt){
// i^=1;
int cap=e[i^1].cap, v=e[i^1].u, flow=e[i^1].flow;
if(!vis[v] && cap>flow){
vis[v]=true;
q.push(v);
d[v]=d[u]+1;
}
// i^=1;
}
}
}
vector<int> Mincut(){
bfs();
vector<int> ans;
for(int i=0; i<cc; i++){
int u=e[i].u, v=e[i].v, cap=e[i].cap;
if(!vis[u] && vis[v] && cap>0) ans.push_back(i);
}
return ans;
}
int num[MAXN], p[MAXN], n;
int Augment(){
int u=t, a=INF;
while(u!=s){
a=MIN(a, e[p[u]].cap-e[p[u]].flow);
u=e[p[u]].u;
// cout<<"a"<<endl;
}
for(u=t; u!=s; u=e[p[u]].u){
e[p[u]].flow+=a;
e[p[u]^1].flow-=a;
// cout<<"b"<<endl;
}
return a;
}
int isap(int need){
bfs();
memset(num, 0, sizeof(num));
int i;
for(i=0; i<n; i++) num[d[i]]++;
for(i=0; i<n; i++) cur[i]=h[i];
int flow=0;
for(int u=s; d[s]<n;){
if(u==t){
flow+=Augment();
if(flow>=need) return flow;
u=s;
}
bool ok=false;
for(i=cur[u]; i!=-1; i=e[i].nxt){
int v=e[i].v, cap=e[i].cap, flow=e[i].flow;
if(d[u] == d[v]+1 && cap>flow){
ok=true;
p[v]=i;
cur[u]=i;
u=v;
break;
}
}
if(!ok){
int tmp=n-1;
for(int i=h[u]; i!=-1; i=e[i].nxt) if(e[i].cap>e[i].flow)
tmp=MIN(tmp, d[e[i].v]);
if(--num[d[u]]==0) break;
num[d[u]=tmp+1]++;
cur[u]=h[u];
if(u!=s) u=e[p[u]].u;
}
// cout<<"adf"<<endl;
}
return flow;
}
void clearFlow(){
for(int i=0; i<cc; i++) e[i].flow=0;
}
int main(){
// freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
int ee, c, T=1;
while(scanf(" %d %d %d", &n, &ee, &c)==3 && n){
int i, u, v, cap;
memset(h, -1, sizeof(h)); cc=0;
for(i=0; i<ee; i++){
scanf(" %d %d %d", &u, &v, &cap); u--; v--;
add(u, v, cap, 0); add(v, u, 0, 0);
}
s=0; t=n-1;
printf("Case %d: ", T++);
int flow=isap(INF);
if(flow>=c){
printf("possible
"); continue;
}
vector<int> mincut=Mincut();
vector<edge> ans;
reduce();
for(int i=0; i<mincut.size(); i++){
int &cap=e[mincut[i]].cap;
cap=c;
clearFlow();
if(flow+isap(c-flow)>=c) ans.push_back(e[mincut[i]]);
cap=0;
}
if(ans.empty()){
printf("not possible
"); continue;
}
sort(ans.begin(), ans.end());
printf("possible option:(%d,%d)", ans[0].u+1, ans[0].v+1);
for(i=1; i<ans.size(); i++) printf(",(%d,%d)", ans[i].u+1, ans[i].v+1);
printf("
");
}
return 0;
}