标签：some   event   iostream   return   problem   sizeof   ret   mmu   stream   

链接：

https://vjudge.net/problem/POJ-1523#author=0

题意：

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.

思路：

dfs求割点。每从一个点扩展出去一个割点，Num就加1，记录一个联通快。
最后答案加1，记录父亲的联通快。

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
using namespace std;
typedef long long LL;
const int MAXN = 1e3+10;

int Num[MAXN];
int First[MAXN];
int To[MAXN], Next[MAXN];
int Dfn[MAXN], Low[MAXN];
int s = 1e3+10, e = 0;
int times, line;

void Dfs(int u, int v)
{
    Dfn[v] = Low[v] = ++times;
    int son = 0;
    for (int i = First[v];i != 0;i = Next[i])
    {
        int node = To[i];
        if (node == u)
            continue;
        if (Dfn[node] == 0)
        {
            son++;
            Dfs(v, node);
            Low[v] = min(Low[v], Low[node]);
            if ((v == s && son >= 2) || (v != s && Low[v] >= Dfn[v]))
                Num[v]++;
        }
        else
            Low[v] = min(Low[v], Dfn[node]);
    }
}

void Init()
{
    memset(Dfn, 0, sizeof(Dfn));
    memset(First, 0, sizeof(First));
    memset(To, 0, sizeof(To));
    memset(Next, 0, sizeof(Next));
    memset(Num, 0, sizeof(Num));
    times = line = 0;
    s = 1e3+10;
}

void AddEdge(int u, int v)
{
    To[++line] = v;
    Next[line] = First[u];
    First[u] = line;
    To[++line] = u;
    Next[line] = First[v];
    First[v] = line;
}

int main()
{
    int l, r;
    int cnt = 0;
    while (~scanf("%d", &l)&& l)
    {
        scanf("%d", &r);
        s = min(s, l);
        s = min(s, r);
        AddEdge(l, r);
        while (~scanf("%d", &l)&& l)
        {
            scanf("%d", &r);
            s = min(s, l);
            s = min(s, r);
            AddEdge(l, r);
        }
        Dfs(0, s);
        printf("Network #%d\n", ++cnt);
        bool flag = true;
        for (int i = 1;i <= 1000;i++)
        {
            if (Num[i])
            {
                printf("  SPF node %d leaves %d subnets\n", i, Num[i]+1);
                flag = false;
            }
        }
        if (flag)
            printf("  No SPF nodes\n");
        printf("\n");
        Init();
    }

    return 0;
}

POJ-1523-SPF（求割点）

标签：some   event   iostream   return   problem   sizeof   ret   mmu   stream   

原文地址：https://www.cnblogs.com/YDDDD/p/11188289.html