zoukankan      html  css  js  c++  java
  • HDU4612 Warm up —— 边双联通分量 + 重边 + 缩点 + 树上最长路

    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
     
    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
    ", 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
    ", block-1-diameter);
    123     }
    124 }
    View Code
     
  • 相关阅读:
    Java基础学习总结(42)——Log4j 2使用教程
    Java基础学习总结(42)——Log4j 2使用教程
    Myeclipse学习总结(1)——Myeclipse优化配置
    Python不再为字符集编码发愁,使用chardet轻松解决你的困扰。
    补习系列(6)- springboot 整合 shiro 一指禅
    给一线讲产品·8期|VPC、子网、安全组,是什么关系?
    Python不再为字符集编码发愁,使用chardet轻松解决你的困扰。
    python推导式pythonic必备
    【nodejs原理&源码赏析(6)】深度剖析cluster模块源码与node.js多进程(下)
    【nodejs原理&源码赏析(5)】net模块与通讯的实现
  • 原文地址:https://www.cnblogs.com/DOLFAMINGO/p/7689173.html
Copyright © 2011-2022 走看看