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2013
11-07

POJ 1523 SPF [解题报告] C++

SPF

问题描述 :

Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible.

Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate.

输入:

The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

输出:

For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist.

The first network in the file should be identified as “Network #1″, the second as “Network #2″, etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text “No SPF nodes” instead of a list of SPF nodes.

样例输入:

1 2
5 4
3 1
3 2
3 4
3 5
0

1 2
2 3
3 4
4 5
5 1
0

1 2
2 3
3 4
4 6
6 3
2 5
5 1
0

0

样例输出:

Network #1
  SPF node 3 leaves 2 subnets

Network #2
  No SPF nodes

Network #3
  SPF node 2 leaves 2 subnets
  SPF node 3 leaves 2 subnets

解题代码:

//* @author: [email protected]
#include 
using namespace std;
int edge[1001][1001];
int used[1001];
int n,cnt,dfn[1001],low[1001],son,sub[1001];
void init()
{
	low[1]=dfn[1]=1;
	cnt=1;
	son=0;
	memset(used,0,sizeof(used));
	used[1]=1;
	memset(sub,0,sizeof(sub));
}

void dfs(int u)
{
 int v;
 for(v=1;v<=n;v++)
 {
  if(!edge[u][v]) continue;
  edge[u][v]=edge[v][u]=0;
  if(!used[v])
  {
	used[v]=1;
	cnt++;
	dfn[v]=low[v]=cnt;
	dfs(v);
	if(low[u]>low[v]) low[u]=low[v];
	if(low[v]>=dfn[u])
	{
	  if(u!=1) sub[u]++;
	  else son++;
	}
   }
   else if(low[u]>dfn[v])low[u]=dfn[v];
  }
 }

 int main()
 {
  int i,u,v,find,num=1;
  while(1)
  {
   scanf("%d",&u);
   if(!u)break;
   memset(edge,0,sizeof(edge));
   n=0;
   scanf("%d",&v);
   if(u>n) n=u;
   if(v>n) n=v;
   edge[u][v]=edge[v][u]=1;
   while(1)
    {
	scanf("%d",&u);
	if(!u) break;
	scanf("%d",&v);
	if(u>n) n=u;
	if(v>n) n=v;
	edge[u][v]=edge[v][u]=1;
    }
    if(num>1)puts("");
    printf("Network #%d\n",num++);
    init();
    dfs(1);
    if(son>1) sub[1]=son-1;
    find=0;
    for(i=1;i<=n;i++)
    {
	if(sub[i])
	{
	  find=1;
	  printf("  SPF node %d leaves %d subnets\n",i,sub[i]+1);
	}
    }
    if(!find) puts("  No SPF nodes");
  }
}

  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。