uva 11248 - Frequency Hopping 最大流最小割入门题 求割集模板

 

At this moment, through out Europe, our base station numbers 1 to N are actively operational through wireless channels. Immediately we require sendingC secret message fragments from our head quarters (base station 1) toNth 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 toColonel 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 numbersN, E andC whereN (0<N<101) represents the number of base stations, E (E<10000) represents the number of available connections between the base stations andC (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) andfp (0<fp<5001) which represent the number of frequency channels available currently fromb1 tob2. 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 sendC 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

 

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=24&problem=2205&mosmsg=Submission+received+with+ID+12258596

 

 

 

刘汝佳新书--训练指南

 

题意：给定一个有向网络，每条边均有一个容量。问是否存在一个从点1到点N，流量为C的流。如果不存在，是否可以恰好修改一条弧的容量，使得存在这样的流？

 

 

 

 

 

分析：先求一次最大流，如果流量至少为C，则直接输出possible，否则需要修改的弧一定是最小割里的弧。依次把这些弧的容量增加到C，然后再求最大流，看最大流量是否至少为C即可。

很可惜，这样写出来的程序会超时，还需要加两个重要的优化。第一个优化是求完最大流后把流量留着，以后每次在它的基础上增广，第二个优化是每次没必要求出最大流，增广到流量至少为C时就停下来。

 

 
 
#include<iostream>  
#include<string>  
#include<algorithm>  
#include<cstdlib>  
#include<cstdio>  
#include<set>  
#include<map>  
#include<vector>  
#include<cstring>  
#include<stack>  
#include<cmath>  
#include<queue>  
using namespace std;  
#define CL(x,v); memset(x,v,sizeof(x));  
#define INF 0x3f3f3f3f  
#define LL long long  
#define MAXN 5000  
struct Edge{  
    int from,to,cap,flow;  
};  
  
struct Dinic{  
    int n,m,s,t;  
    vector<Edge> edges;  
    vector<int> G[MAXN];// 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号  
    bool vis[MAXN];  
    int d[MAXN];  
    int cur[MAXN];  
    void init(int n){  
        this->n=n;  
        for(int i=0;i<=n;i++)G[i].clear();  
        edges.clear();  
    }  
    void AddEdge(int from,int to,int cap){  
        Edge e;  
        e.from=from;e.to=to;e.cap=cap;e.flow=0;  
        edges.push_back(e);  
        e.from=from;e.to=to;e.cap=0;e.flow=0;  
        edges.push_back(e); m=edges.size();  
        G[from].push_back(m-2);  
        G[to].push_back(m-1);  
    }  
    bool BFS(){  
        CL(vis,0);  
        queue<int> Q;  
        Q.push(s);  
        d[s]=0;  
        vis[s]=1;  
        while(!Q.empty()){  
            int x=Q.front();  
            Q.pop();  
            for(int i=0;i<G[x].size();i++){  
                Edge& e=edges[G[x][i]];  
                if(!vis[e.to]&&e.cap>e.flow){  
                    vis[e.to]=1;  
                    d[e.to]=d[x]+1;  
                    Q.push(e.to);  
                }  
            }  
        }  
        return vis[t];  
    }  
    int DFS(int x,int a){  
        if(x==t||a==0)return a;  
        int flow=0,f;
补充：软件开发 , C++ ,