zoukankan      html  css  js  c++  java
  • POJ 3352 Road Construction

    It's almost summer time, and that means that it's almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

    The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

    Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

    So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

    Input

    The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

    Output

    One line, consisting of an integer, which gives the minimum number of roads that we need to add.

    Sample Input
    Sample Input 1
    10 12
    1 2
    1 3
    1 4
    2 5
    2 6
    5 6
    3 7
    3 8
    7 8
    4 9
    4 10
    9 10
    
    Sample Input 2
    3 3
    1 2
    2 3
    1 3
    Sample Output
    Output for Sample Input 1
    2
    
    Output for Sample Input 2
    0
    

     

    与 POJ 3117 差不多,只是两点之间边是唯一的。与 POJ 3117 差不多,只是两点之间边是唯一的。直接用 POJ 3117 代码可AC

     

     

    #include <map>
    #include <set>
    #include <cmath>
    #include <ctime>
    #include <queue>
    #include <stack>
    #include <cstdio>
    #include <string>
    #include <vector>
    #include <cstdlib>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    #define ll long long
    #define file(a) freopen(a".in","r",stdin); freopen(a".out","w",stdout);
    
    inline int gi()
    {
        bool b=0; int r=0; char c=getchar();
        while(c<'0' || c>'9') { if(c=='-') b=!b; c=getchar(); }
        while(c>='0' && c<='9') { r=r*10+c-'0'; c=getchar(); }
        if(b) return -r; return r;
    }
    
    const int inf = 1e9+7, N = 1007;
    int n,m,num,Deep,f[N],dfn[N],low[N],cd[N];
    bool b[N];
    stack <int> s;
    struct data
    {
        int fr,to,nx;
    }da[N];
    
    inline void add (int fr,int to)
    {
        da[++num].fr=fr, da[num].to=to, da[num].nx=f[fr], f[fr]=num;
    }
    
    inline void tarjan (int o,int fa)
    {
        dfn[o]=low[o]=++Deep; b[o]=1;
        int i,to;
        for (i=f[o]; i; i=da[i].nx)
            {
                to=da[i].to;
                if (to == fa) continue;
                if (!dfn[to])
                    {
                        tarjan (to,o);
                        low[o]=min(low[o],low[to]);
                    }
                else if (b[to]) low[o]=min(low[o],dfn[to]);
            }
        b[o]=0;
    }
    
    int main()
    {
    //    file("POJ-3352");
        n=gi(), m=gi();
        int i,j,x,y;
        for (i=1; i<=m; i++)
            {
                x=gi(), y=gi();
                add (x,y), add(y,x);
            }
        for (i=1; i<=n; i++) if (!dfn[i]) tarjan (i,0);
        for (i=1; i<=n; i++)
            for (j=f[i]; j; j=da[j].nx)
                {
                    x=low[da[j].to]; y=low[i];
                    if (y != x) cd[y]++;
                }
        x=0;
        for (i=1; i<=n; i++) if (cd[i] == 1) x++;
        printf("%d
    ",(x+1)/2);
        return 0;
    }

     

     

    欢迎在评论区提问质疑!

     

  • 相关阅读:
    破衣服的回忆
    underscorejs 源码走读笔记
    关于书籍《区块链以太坊DApp开发实战》的内容告示
    从区块链技术研发者的角度,说说我的区块链从业经历和对它的理解
    简介 以太坊 2.0 核心 之 共识机制的改变
    一般电商应用的订单队列架构思想
    详细讲解:零知识证明 之 ZCash 完整的匿名交易流程
    HyperLogLog 算法的原理讲解以及 Redis 是如何应用它的
    由 System.arraycopy 引发的巩固:对象引用 与 对象 的区别
    如何独立开发一个网络请求框架
  • 原文地址:https://www.cnblogs.com/y142857/p/6883270.html
Copyright © 2011-2022 走看看