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  • POJ 1523 SPF

    SPF

    Time Limit: 1000ms
    Memory Limit: 10000KB
    This problem will be judged on PKU. Original ID: 1523
    64-bit integer IO format: %lld      Java class name: Main
     
    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. 
     

    Input

    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.
     

    Output

    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.
     

    Sample Input

    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

    Sample Output

    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

    Source

    Greater New York 2000
     
    解题：求割点。然后dfs计算可以分成多少个块。
     
     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 1010;
18 vector<int>g[maxn];
19 bool iscut[maxn];
20 int dfn[maxn],low[maxn],cnt,vis[maxn];
21 void tarjan(int u,int fa) {
22     dfn[u] = low[u] = ++cnt;
23     vis[u] = 1;
24     int son = 0;
25     for(int i = 0; i < g[u].size(); i++) {
26         if(!vis[g[u][i]]) {
27             tarjan(g[u][i],u);
28             son++;
29             low[u] = min(low[u],low[g[u][i]]);
30             if(fa == -1 && son > 1 || fa != -1 && low[g[u][i]] >= dfn[u])
31                 iscut[u] = true;
32         } else if(vis[g[u][i]] == 1) low[u] = min(low[u],dfn[g[u][i]]);
33     }
34     vis[u] = 2;
35 }
36 void dfs(int u) {
37     vis[u] = 1;
38     for(int i = 0; i < g[u].size(); i++)
39         if(!vis[g[u][i]]) dfs(g[u][i]);
40 }
41 void init() {
42     memset(vis,0,sizeof(vis));
43     memset(iscut,false,sizeof(iscut));
44     for(int i = 0; i < maxn; i++) g[i].clear();
45 }
46 int main() {
47     int i,j,u,v,ks = 1,n,son;
48     bool flag;
49     while(scanf("%d",&u),u) {
50         n = 0;
51         init();
52         scanf("%d",&v);
53         n = max(max(u,v),n);
54         g[u].push_back(v);
55         g[v].push_back(u);
56         while(scanf("%d",&u),u) {
57             scanf("%d",&v);
58             n = max(max(u,v),n);
59             g[u].push_back(v);
60             g[v].push_back(u);
61         }
62         flag = true;
63         tarjan(1,-1);
64         printf("Network #%d
",ks++);
65         for(i = 1; i <= n; i++) {
66             if(iscut[i]) {
67                 flag = false;
68                 son = 0;
69                 memset(vis,0,sizeof(vis));
70                 vis[i] = 1;//记得把路锁死
71                 for(j = 0; j < g[i].size(); j++) {
72                     if(!vis[g[i][j]]) {
73                         son++;
74                         dfs(g[i][j]);
75                     }
76                 }
77                 printf("  SPF node %d leaves %d subnets
",i,son);
78             }
79         }
80         if(flag) puts("  No SPF nodes");
81         puts("");
82     }
83     return 0;
84 }
    View Code

    不用再次dfs的方案：

    去掉该关节点u，将原来的连通图分成几个连通分量？答案如下：

    (1)如果关节点u是根结点，则有几个子女，就分成几个连通分量。

    (2)如果关节点u不是根节点，则有n个子女w，使得low[w]>= dfn[u]，则去掉该点，分成了n+1个连通分量。

    注意：第二条的儿子数不同于第一个条件的儿子数。。。第二个多了个约束

     1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 1010;
 7 struct arc {
 8     int to,next;
 9     arc(int x = 0,int y = -1) {
10         to = x;
11         next = y;
12     }
13 };
14 arc e[maxn*100];
15 int head[maxn],dfn[maxn],low[maxn],cut[maxn],tot,scc,idx;
16 void add(int u,int v) {
17     e[tot] = arc(v,head[u]);
18     head[u] = tot++;
19 }
20 void tarjan(int u,int fa) {
21     dfn[u] = low[u] = idx++;
22     int son = 0,gson = 0;
23     bool flag = true,iscut = false;
24     for(int i = head[u]; ~i; i = e[i].next) {
25         if(flag && e[i].to == fa) {
26             flag = false;
27             continue;
28         }
29         if(dfn[e[i].to] == -1) {
30             tarjan(e[i].to,u);
31             son++;
32             low[u] = min(low[u],low[e[i].to]);
33             if(fa == -1 && son >= 2) iscut = true;
34             if(fa != -1 && low[e[i].to]>= dfn[u]) iscut = true,gson++;
35             } else if(e[i].to != fa) low[u] = min(low[u],dfn[e[i].to]);
36     }
37     if(iscut && fa == -1) cut[u] = son-1;
38     else if(iscut && fa != -1) cut[u] = gson;
39 }
40 int main() {
41     int u,v,kase = 1;
42     while(scanf("%d",&u),u) {
43         scanf("%d",&v);
44         memset(head,-1,sizeof(head));
45         memset(cut,0,sizeof(cut));
46         memset(dfn,-1,sizeof(dfn));
47         memset(low,-1,sizeof(low));
48         idx = tot = 0;
49         add(u,v);
50         add(v,u);
51         int up = max(u,v);
52         while(scanf("%d",&u),u) {
53             scanf("%d",&v);
54             add(u,v);
55             add(v,u);
56             up = max(up,max(u,v));
57         }
58         for(int i = 1; i <= up; ++i)
59             if(dfn[i] == -1) tarjan(i,-1);
60         int ans = 0;
61         printf("Network #%d
",kase++);
62         for(int i = 1; i <= up; ++i) {
63             if(cut[i]) {
64                 printf("  SPF node %d leaves %d subnets
",i,cut[i]+1);
65                 ans++;
66             }
67         }
68         if(!ans) puts("  No SPF nodes");
69         puts("");
70     }
71     return 0;
72 }
    View Code
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  • 原文地址：https://www.cnblogs.com/crackpotisback/p/3939372.html
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