POJ 3177--Redundant Paths【无向图添加最少的边成为边双连通图 && tarjan求ebc && 缩点构造缩点树】

Redundant Paths

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10798		Accepted: 4626

Description

In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular
path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only
travel on Official Paths when they move from one field to another. 



Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate
routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way. 



There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

Input

Line 1: Two space-separated integers: F and R 



Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

Output

Line 1: A single integer that is the number of new paths that must be built.

Sample Input

7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7

Sample Output

2

Hint

Explanation of the sample: 



One visualization of the paths is:

   1   2   3
   +---+---+
       |   |
       |   |
 6 +---+---+ 4
      / 5
     /
    /
 7 +

Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.

   1   2   3
   +---+---+
   :   |   |
   :   |   |
 6 +---+---+ 4
      / 5  :
     /     :
    /      :
 7 + - - - - 

Check some of the routes: 

1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2 

1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4 

3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7 

Every pair of fields is, in fact, connected by two routes. 



It's possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.

下面解析来自：女神的博客

斌神博客上有个不错的总结：斌神的博客

大致题意：

为了保护放牧环境，避免牲畜过度啃咬同一个地方的草皮，牧场主决定利用不断迁移牲畜进行喂养的方法去保护牧草。然而牲畜在迁移过程中也会啃食路上的牧草，所以假设每次迁移都用同一条道路，那么该条道路相同会被啃咬过度而遭受破坏。

如今牧场主拥有F个农场。已知这些农场至少有一条路径连接起来（不一定是直接相连）。但从某些农场去另外一些农场。至少有一条路可通行。为了保护道路上的牧草，农场主希望再建造若干条道路，使得每次迁移牲畜时，至少有2种迁移途径，避免反复走上次迁移的道路。

已知当前有的R条道路。问农场主至少要新建造几条道路，才干满足要求？

解题思路：

“使得每次迁移牲畜时，至少有2种迁移途径，避免反复走上次迁移的道路。”就是说当吧F个农场看作点、路看作边构造一个无向图G时，图G不存在桥。

那么能够建立模型：

给定一个连通的无向图G，至少要加入几条边。才干使其变为双连通图。

当图G存在桥（割边）的时候，它必然不是双连通的。桥的两个端点必然分别属于图G的两个【边双连通分量】。一旦删除了桥，这两个【边双连通分量】必然断开，图G就不连通了。可是假设在两个【边双连通分量】之间再加入一条边。桥就不再是桥了。这两个【边双连通分量】之间也就是双连通了。



那么假设图G有多个【边双连通分量】呢？至少应该加入多少条边，才干使得随意两个【边双连通分量】之间都是双连通（也就是图G是双连通的）

1、 首先要找出图G的全部【边双连通分量】。

2、 把每个【边双连通分量】都看做一个点（即【缩点】）

3、 问题再次被转化为“至少在缩点树上添加多少条树边。使得这棵树变为一个双连通图”。

首先知道一条等式：

若要使得随意一棵树。在添加若干条边后。变成一个双连通图，那么

至少添加的边数 =（ 这棵树总度数为1的结点数 + 1 ）/ 2。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#define maxn 5010
#define maxm 20010
using namespace std;

int n, m;
struct node {
    int u, v, next;
};
node edge[maxm];
//缩点后形成树，每一个点的度数
int du[maxn];
int head[maxn], cnt;
int low[maxn], dfn[maxn];
//Belong数组的值是 1 ~ ebc_block
int Stack[maxn], Belong[maxn];
int ebc_block;//边双连通块数
int dfs_clock;
int top;//模拟栈的指针
bool Instack[maxn];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v){
    edge[cnt] = {u, v, head[u]};
    head[u] = cnt++;
}

void getmap(){
    while(m--){
        int a, b;
        scanf("%d%d", &a, &b);
        addedge(a, b);
        addedge(b, a);
    }
}

void tarjan(int u, int pre){
    int v;
    low[u] = dfn[u] = ++dfs_clock;
    Stack[top++] = u;
    Instack[u] = true;
    int have = 1;
    for(int i = head[u]; i != -1; i = edge[i].next){
        v = edge[i].v;
        if(have && v == pre){//去重边
        	have = 0;
        	continue;
		}
        if(!dfn[v]){
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(Instack[v])
            low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u]){
        ebc_block++;
        do{
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = ebc_block;
        }
        while(v != u);
    }
}

void suodian(){
    memset(du, 0, sizeof(du));
    for(int i = 0; i < cnt; i += 2 ){
        int u = Belong[edge[i].u];
        int v = Belong[edge[i].v];
        if(u != v)
            du[u]++, du[v]++;
    }
}

void find(){
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(Instack, false, sizeof(Instack));
    memset(Belong, 0, sizeof(Belong));
    dfs_clock = 0;
    ebc_block = 0;
    top = 0;
    tarjan(1, -1);//连通图
}

void solve(){
    int ans = 0;
    if(ebc_block == 1){
        printf("0\n");
        return ;
    }
    for(int i = 1; i <= ebc_block; ++i)
        if(du[i] == 1) ans++;
    printf("%d\n", (ans + 1) / 2);
}

int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        init();
        getmap();
        find();
        suodian();
        solve();
    }
    return 0;
}