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

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

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