解题报告 之 UVA820 Internet Bandwidth

Description

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.

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.

题目大意：有一个计算机网络，输入节点数n，输入网络流源点和汇点src，des，再输入双向边数m。给出m条边的负载，求最大流。

分析：裸的最大流，连背景都不包装了，于是乎直接敲了模板跑最大流即可。注意唯一的坑点在于输出要多一行空行，没看到就一直WA把。于是乎解题报告至此结束啦。

上代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

const int MAXM = 160000;
const int MAXN = 400;
const int INF = 0x3f3f3f3f;

struct Edge
{
	int to, cap, next;
};

Edge edge[MAXM];
int level[MAXN];
int head[MAXN];
int src, des, cnt;

void addedge( int from, int to, int cap )
{
	edge[cnt].to = to;
	edge[cnt].cap = cap;
	edge[cnt].next = head[from];
	head[from] = cnt++;

	swap( from, to );

	edge[cnt].to = to;
	edge[cnt].cap = cap;
	edge[cnt].next = head[from];
	head[from] = cnt++;
}

int bfs()
{
	memset( level, -1, sizeof level );
	queue<int>q;
	while (!q.empty())
		q.pop();

	level[src] = 0;
	q.push( src );

	while (!q.empty())
	{
		int u = q.front();
		q.pop();

		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			if (edge[i].cap > 0 && level[v] == -1)
			{
				level[v] = level[u] + 1;
				q.push( v );
			}
		}
	}
	return level[des] != -1;
}

int dfs( int u, int f )
{
	if (u == des) return f;
	int tem;
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if (edge[i].cap>0&&level[v] == level[u] + 1)
		{
			tem = dfs( v, min( f, edge[i].cap ) );
			if (tem > 0)
			{
				edge[i].cap -= tem;
				edge[i^1].cap += tem;
				return tem;
			}
		}
	}
	level[u] = -1;
	return 0;
}

int Dinic()
{
	int ans = 0, tem;

	while (bfs())
	{
		while ((tem = dfs( src, INF )) > 0)
		{
			ans += tem;
		}
	}
	return ans;
}

int main()
{
	int n, m;
	int kase = 1;
	while (cin >> n&&n)
	{
		memset( head, -1, sizeof head );
		cnt = 0;
		cin >> src >> des >> m;
		for (int i = 1; i <= m; i++)
		{
			int a, b, c;
			cin >> a >> b >> c;
			addedge( a, b, c );
		}

		printf( "Network %d\nThe bandwidth is %d.\n\n",kase++, Dinic() );
	}
	return 0;

}

啦啦啦啦啦啦啦，好有诚意的一道题。。