uva 11248 最大流 ISAP

E	Frequency Hopping
Input: Standard Input Output: Standard Output

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

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

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

Case 1: possible Case 2: possible option:(1,2),(3,4) Case 3: not possible

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\n");   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\n");   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("\n");
}
return 0;
}