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

Author

SYSU

Source

2013 Multi-University Training Contest 2

Recommend

zhuyuanchen520

題解：

1.用Tarjan算法求出每個邊雙聯通分量，由于每一對點之間可以有多條邊，所以在判斷邊是否被重復訪問時，需要依據邊的下標而定?。

2.對每個邊雙聯通分量進行縮點，縮點之后得到的是一棵無根樹。

3.在樹上添加一條邊，使得橋的數目減少最多。最多能減少多少呢？樹上最長路。

vector建樹：

  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         //因為一對點之間可能有多條邊，所以不能根據v是否為上一個點來防止邊是否被重復訪問。而需要根據邊的編號
 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

前向星建樹：

  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         //因為一對點之間可能有多條邊，所以不能根據v是否為上一個點來防止邊是否被重復訪問。而需要根據邊的編號
 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

轉載于:https://www.cnblogs.com/DOLFAMINGO/p/7689173.html

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