poj 1523(无向联通图的割点)
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poj 1523(无向联通图的割点)
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結合tarjan算法思想，這題終于寫了出來。
同樣用dfs將圖變成為一顆樹，這樣可以提供許多有用的性質。
對于一個無向連通圖，dfs后的樹為只有回邊（回邊Euv,v是u的祖先）和生成樹的邊的圖。 那么在遍歷到一個點u的時候，可以知道如果不考慮這個點，如果與u相鄰的點連通那么u不是割點，否則是割點。 那么只需要判斷與u相鄰的點是否連通就行了，于是借鑒tarjan求強連通的辦法，在dfs時，對每個點標記一個深度low[N]也就是從根到這個點最短路徑（經過的最小結點數）， 然后在遍歷到u點的時候，看看與u相鄰的點v的low[v], 如果low[U] >= low[u]那么說明u就是割點. ?因為v點無法到達u相鄰的某些點。 當然當u為根節點的時候要特判
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SPF
Description 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 01 2 2 3 3 4 4 5 5 1 01 2 2 3 3 4 4 6 6 3 2 5 5 1 00 Sample Output Network #1SPF node 3 leaves 2 subnetsNetwork #2No SPF nodesNetwork #3SPF node 2 leaves 2 subnetsSPF node 3 leaves 2 subnets Source Greater New York 2000 |
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// start time 12:44 #include <iostream> #include <string.h> #include <stdio.h> using namespace std; #define N 1010 #define INF 0x3ffffff int g[N][N]; int mark[N]; int low[N]; int save[N]; int n;void dfs(int s,int cnt) {low[s]=cnt;int mi=INF;int k=0;for(int i=1;i<=n;i++){if(g[s][i]==0||s==i) continue;if(low[i] == -1){dfs(i,cnt+1);if( low[i] >= cnt ) k++;}mi=min(low[i],mi);}low[s]=mi;if(cnt==0) k--;if(k!=0) save[s]=k; }int main() {int tt=1;int x,y;while(scanf("%d",&x)&&x){memset(mark,0,sizeof(mark));mark[x]=1;n=0;if(x>n) n=x;memset(g,0,sizeof(g));memset(save,-1,sizeof(save));memset(low,-1,sizeof(low));scanf("%d",&y);mark[y]=1;if(y>n) n=y;g[x][y]=g[y][x]=1;while(scanf("%d",&x)&&x){mark[x]=1;if(x>n) n=x;scanf("%d",&y);mark[y]=1;if(y>n) n=y;g[x][y]=g[y][x]=1;}for(int i=1;i<=n;i++)if(mark[i]==1){dfs(i,0);break;}printf("Network #%d\n",tt++);int flag=0;for(int i=1;i<=n;i++){if(save[i]!=-1){flag=1;printf(" SPF node %d leaves %d subnets\n",i,save[i]+1);}}if(flag==0)printf(" No SPF nodes\n");printf("\n");}return 0; }
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轉載于:https://www.cnblogs.com/chenhuan001/archive/2013/05/13/3075662.html
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