UVA 820 Internet Bandwidth

 

  Internet Bandwidth 

On the Internet, machines (nodes) are richly interconnected, and many paths may exist between a given pair of nodes. The total message-carrying capacity (bandwidth) between two given nodes is the maximal amount of data per unit time that can be transmitted from one node to the other. Using a technique called packet switching, this data can be transmitted along several paths at the same time.

For example, the following figure shows a network with four nodes (shown as circles), with a total of five connections among them. Every connection is labeled with a bandwidth that represents its data-carrying capacity per unit time.

In our example, the bandwidth between node 1 and node 4 is 25, which might be thought of as the sum of the bandwidths 10 along the path 1-2-4, 10 along the path 1-3-4, and 5 along the path 1-2-3-4. No other combination of paths between nodes 1 and 4 provides a larger bandwidth.

You must write a program that computes the bandwidth between two given nodes in a network, given the individual bandwidths of all the connections in the network. In this problem, assume that the bandwidth of a connection is always the same in both directions (which is not necessarily true in the real world).

Input 

The input file contains descriptions of several networks. Every description starts with a line containing a single integer n (2 ≤n ≤100), which is the number of nodes in the network. The nodes are numbered from 1 to n. The next line contains three numbers s, t, and c. The numbers s and t are the source and destination nodes, and the number c is the total number of connections in the network. Following this are c lines describing the connections. Each of these lines contains three integers: the first two are the numbers of the connected nodes, and the third number is the bandwidth of the connection. The bandwidth is a non-negative number not greater than 1000.

There might be more than one connection between a pair of nodes, but a node cannot be connected to itself. All connections are bi-directional, i.e. data can be transmitted in both directions along a connection, but the sum of the amount of data transmitted in both directions must be less than the bandwidth.

A line containing the number 0 follows the last network description, and terminates the input.

Output 

For each network description, first print the number of the network. Then print the total bandwidth between the source node s and the destination node t, following the format of the sample output. Print a blank line after each test case.

Sample Input	Output for the Sample Input
4 1 4 5 1 2 20 1 3 10 2 3 5 2 4 10 3 4 20 0	Network 1 The bandwidth is 25.

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ACM World Finals 2000, Problem E
//uva 820
//网络流模板题
//注意题目要求的格式

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <queue>
#define N 103
using namespace std;
int flow[N][N],cap[N][N];
int f[N];
int s,t,n;
int BFS()
{   int a[N];
    int u,v;
    memset(flow,0,sizeof(flow));
    int fl=0;
    for(;;)
    {
        memset(a,0,sizeof(a));
        queue<int>Q;
        a[s]=1003;
        Q.push(s);
        while(!Q.empty())
        {
            u=Q.front();Q.pop();
            for(v=1;v<=n;v++)
            if(!a[v]&&cap[u][v]-flow[u][v]>0)
            {
                f[v]=u;
                a[v]=a[u]>cap[u][v]-flow[u][v]?cap[u][v]-flow[u][v]:a[u];
                Q.push(v);
            }
        }
        if(a[t]==0) break;
        for(u=t;u!=s;u=f[u])
        {
            flow[f[u]][u]+=a[t];
            flow[u][f[u]]-=a[t];
        }
        fl+=a[t];
    }
    return fl;
}
int main()
{
    int c,u,v,cost;
    int tt=1;
    while(scanf("%d",&n),n)
    {
        scanf("%d%d%d",&s,&t,&c);
        memset(cap,0,sizeof(cap));
        while(c--)
        {
          scanf("%d%d%d",&u,&v,&cost);
          cap[u][v]+=cost;
          cap[v][u]=cap[u][v];
        }
        printf("Network %d\n",tt++);
        printf("The bandwidth is %d.\n\n",BFS());
    }
    return 0;
}