题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=4612

 

Warm up

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 7206    Accepted Submission(s): 1681

Problem Description N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
　　If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
　　Note that there could be more than one channel between two planets.

 

Input The input contains multiple cases.
　　Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
　　(2<=N<=200000, 1<=M<=1000000)
　　Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
　　A line with two integers '0' terminates the input.

 

Output For each case, output the minimal number of bridges after building a new channel in a line.

 

Sample Input 4 4 1 2 1 3 1 4 2 3 0 0

 

Sample Output 0

 

Author SYSU

 

Source 2013 Multi-University Training Contest 2

 

Recommend zhuyuanchen520 题解： 1.用Tarjan算法求出每个边双联通分量，由于每一对点之间可以有多条边，所以在判断边是否被重复访问时，需要依据边的下标而定 。 2.对每个边双联通分量进行缩点，缩点之后得到的是一棵无根树。 3.在树上添加一条边，使得桥的数目减少最多。最多能减少多少呢？树上最长路。 vector建树： 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const double EPS = 1e-8; 15 const int INF = 2e9; 16 const LL LNF = 2e18; 17 const int MAXN = 2e5+10; 18 19 struct Edge 20 { 21 int to, next; 22 }edge[MAXN*10]; 23 int tot, head[MAXN]; 24 vector<int>g[MAXN]; 25 26 void addedge(int u, int v) 27 { 28 edge[tot].to = v; 29 edge[tot].next = head[u]; 30 head[u] = tot++; 31 } 32 33 int index, dfn[MAXN], low[MAXN]; 34 int top, Stack[MAXN], instack[MAXN]; 35 int block, belong[MAXN]; 36 37 void Tarjan(int u, int pre) 38 { 39 dfn[u] = low[u] = ++index; 40 Stack[top++] = u; 41 instack[u] = true; 42 for(int i = head[u]; i!=-1; i = edge[i].next) 43 { 44 //因为一对点之间可能有多条边，所以不能根据v是否为上一个点来防止边是否被重复访问。而需要根据边的编号 45 if((i^1)==pre) continue; 46 int v = edge[i].to; 47 if(!dfn[v]) 48 { 49 Tarjan(v, i); 50 low[u] = min(low[u], low[v]); 51 } 52 else if(instack[v]) 53 low[u] = min(low[u], dfn[v]); 54 } 55 56 if(low[u]==dfn[u]) 57 { 58 block++; 59 int v; 60 do 61 { 62 v = Stack[--top]; 63 instack[v] = false; 64 belong[v] = block; 65 }while(v!=u); 66 } 67 } 68 69 int diameter, endpoint; 70 int dfs(int u, int pre, int dep) 71 { 72 if(dep>diameter) { endpoint = u; diameter = dep; } 73 for(int i = 0; i<g[u].size(); i++) 74 if(g[u][i]!=pre) 75 dfs(g[u][i], u, dep+1); 76 } 77 78 void init(int n) 79 { 80 tot = 0; 81 memset(head, -1, sizeof(head)); 82 83 index = 0; 84 memset(dfn, 0, sizeof(dfn)); 85 memset(low, 0, sizeof(low)); 86 87 top = 0; 88 memset(instack, false, sizeof(instack)); 89 90 block = 0; 91 for(int i = 1; i<=n; i++) 92 belong[i] = i, g[i].clear(); 93 } 94 95 int main() 96 { 97 int n, m; 98 while(scanf("%d%d", &n, &m) && (n||m) ) 99 { 100 init(n); 101 for(int i = 1; i<=m; i++) 102 { 103 int u, v; 104 scanf("%d%d", &u, &v); 105 addedge(u, v); 106 addedge(v, u); 107 } 108 109 Tarjan(1, -1); 110 for(int u = 1; u<=n; u++) 111 for(int i = head[u]; i!=-1; i = edge[i].next) 112 { 113 int v = edge[i].to; 114 if(belong[u]!=belong[v]) 115 g[belong[u]].push_back(belong[v]); 116 } 117 118 endpoint = 1, diameter = 0; 119 dfs(1, -1, 0); 120 dfs(endpoint, -1, 0); 121 printf("%d\n", block-1-diameter); 122 } 123 } View Code

 

前向星建树： 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const double EPS = 1e-8; 15 const int INF = 2e9; 16 const LL LNF = 2e18; 17 const int MAXN = 2e5+10; 18 19 struct Edge 20 { 21 int from, to, next; 22 }edge[MAXN*10]; 23 int tot, head[MAXN]; 24 25 void addedge(int u, int v) 26 { 27 edge[tot].from = u; 28 edge[tot].to = v; 29 edge[tot].next = head[u]; 30 head[u] = tot++; 31 } 32 33 int index, dfn[MAXN], low[MAXN]; 34 int top, Stack[MAXN], instack[MAXN]; 35 int block, belong[MAXN]; 36 37 void Tarjan(int u, int pre) 38 { 39 dfn[u] = low[u] = ++index; 40 Stack[top++] = u; 41 instack[u] = true; 42 for(int i = head[u]; i!=-1; i = edge[i].next) 43 { 44 //因为一对点之间可能有多条边，所以不能根据v是否为上一个点来防止边是否被重复访问。而需要根据边的编号 45 if((i^1)==pre) continue; 46 int v = edge[i].to; 47 if(!dfn[v]) 48 { 49 Tarjan(v, i); 50 low[u] = min(low[u], low[v]); 51 } 52 else if(instack[v]) 53 low[u] = min(low[u], dfn[v]); 54 } 55 56 if(low[u]==dfn[u]) 57 { 58 block++; 59 int v; 60 do 61 { 62 v = Stack[--top]; 63 instack[v] = false; 64 belong[v] = block; 65 }while(v!=u); 66 } 67 } 68 69 int diameter, endpoint; 70 int dfs(int u, int pre, int dep) 71 { 72 if(dep>diameter) { endpoint = u; diameter = dep; } 73 for(int i = head[u]; i!=-1; i = edge[i].next) 74 if(edge[i].to!=pre) 75 dfs(edge[i].to, u, dep+1); 76 } 77 78 void init(int n) 79 { 80 tot = 0; 81 memset(head, -1, sizeof(head)); 82 83 index = 0; 84 memset(dfn, 0, sizeof(dfn)); 85 memset(low, 0, sizeof(low)); 86 87 top = 0; 88 memset(instack, false, sizeof(instack)); 89 90 block = 0; 91 for(int i = 1; i<=n; i++) 92 belong[i] = i; 93 } 94 95 int main() 96 { 97 int n, m; 98 while(scanf("%d%d", &n, &m) && (n||m) ) 99 { 100 init(n); 101 for(int i = 1; i<=m; i++) 102 { 103 int u, v; 104 scanf("%d%d", &u, &v); 105 addedge(u, v); 106 addedge(v, u); 107 } 108 109 Tarjan(1, -1); 110 tot = 0; 111 memset(head, -1, sizeof(head)); 112 for(int i = 0; i<2*m; i++) 113 { 114 int u = edge[i].from, v = edge[i].to; 115 if(belong[u]!=belong[v]) 116 addedge(belong[u], belong[v]); 117 } 118 119 endpoint = 1, diameter = 0; 120 dfs(1, -1, 0); 121 dfs(endpoint, -1, 0); 122 printf("%d\n", block-1-diameter); 123 } 124 } View Code

 

 

 

 

转载于:https://www.cnblogs.com/DOLFAMINGO/p/7689173.html

总结

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