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.
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
题目大意：直接抽象一下问题：给你一个无向连通图，计算最少需要添加多少边，才能使得任意两点之间至少有两条相互“边独立”的道路（即两条路径中没有相同的边）。
解题思路：这里要用到边双连通分量的知识，先解释一下：在边双连通分量中，不存在割边，其中任何一对顶点之间至少存在两条无公共边的路径（允许有公共内部顶点）。很容易看出，边连通分量中的所有点可以缩为一个点。这样原图就大大简化了，缩点之后的图中的边就只剩下桥了，然后统计出新生成的图（准确说应该是树）中的度为 1 的顶点个数sum ，运用结论（sum + 1）/ 2 就得到答案了。
Ps：缩点时要用到并查集。。
请看代码：
#include<iostream> #include<cstring> #include<string> #include<cmath> #include<cstdio> #include<vector> #include<algorithm> using namespace std ; const int MAXN = 5005 ; struct Node { int adj ; int e ; // 边的序号 Node *next ; } ; Node mem[MAXN * 2] ; // 边节点的数组 int memp ; // 统计边节点 Node *vert[MAXN] ; // 顶点指针数组 int set[MAXN] ; // 用于并查集 bool vis[MAXN] ; // 标记数组，记录顶点是否被访问过 bool vise[MAXN * 2] ; // 标记数组，记录边是否被访问过 int low[MAXN] ; // 记录顶点在深度优先搜索生成树中通过自己的子孙（如果有的话）以及一条回边 // 可以到达的最小深度 int dfn[MAXN] ; // 记录顶点在深度优先搜索生成树中所在的深度 int bridges[MAXN * 2][2] ; // 记录桥的两端顶点 int belong[MAXN] ; // 记录每个顶点所属的边连通分量 int d[MAXN] ; // 统计桥的两端顶点（缩点之后）的度 int cbridges ; // 记录原图中桥的个数 int tmpdfn ; int counte ; // 给图中的边编号 int sumfz ; // 统计原图中边连通分量个数 int root ; // 记录根节点 int n , m ; void clr() // 初始化 { memp = 0 ; counte = 0 ; memset(vis , 0 ,sizeof(vis)) ; memset(vert , 0 , sizeof(vert)) ; memset(vise , 0 , sizeof(vise)) ; memset(belong , -1 , sizeof(belong)) ; memset(bridges , -1 , sizeof(bridges)) ; memset(low , 0 , sizeof(low)) ; memset(dfn , 0 , sizeof(dfn)) ; } int find(int x) // 并查集（查找部分） { int r = x ; int t ; while (r != set[r]) { r = set[r] ; } /* while (x != set[x]) // 并查集的优化 , 可以加在程序中 { t = set[x] ; set[x] = r ; x = t ; }*/ return r ; } void unitset(int i , int j) // 并查集（合并部分） { int tx = find(i) ; int ty = find(j) ; if(tx < ty) { set[ty] = tx ; } else { set[tx] = ty ; } } void init() // 输入 { int i , j ; for(i = 0 ; i < m ; i ++) { int a , b ; scanf("%d%d" , &a , &b) ; //建图 mem[memp].adj = b ; mem[memp].e = counte ; mem[memp].next = vert[a] ; vert[a] = &mem[memp] ; memp ++ ; mem[memp].adj = a ; mem[memp].e = counte ++ ; mem[memp].next = vert[b] ; vert[b] = &mem[memp] ; memp ++ ; root = b ; } } void dfs(int u) // 找桥 { Node *p = vert[u] ; while (p != NULL) { int v = p -> adj ; int te = p -> e ; if(!vise[te]) // 图中可能有重边，所以应先判断此边是否被访问过 { vise[te] = 1 ; if(!vis[v]) { vis[v] = 1 ; dfn[v] = low[v] = ++ tmpdfn ; dfs(v) ; low[u] = min(low[u] , low[v]) ; if(low[v] > dfn[u]) // (u , v) 是桥 { bridges[cbridges][0] = u ; bridges[cbridges ++][1] = v ; } else // 如果（u，v）不是桥，那么u、v必在一个边连通分量中 { unitset(u , v) ; } } else { low[u] = min(low[u] , dfn[v]) ; } } p = p -> next ; } } int countfz() //统计边连通分支数（缩点） { int i ; int k ; int fz = 0 ; for(i = 1 ; i <= n ; i ++) { k = find(i) ; if(belong[k] == -1) { belong[k] = fz ++ ; } belong[i] = belong[k] ; // 缩点 } return fz ; } void solve() { int i ; for(i = 1 ; i <= n ; i ++) // 初始化并查集 { set[i] = i ; } tmpdfn = 1 ; cbridges = 0 ; vis[root] = 1 ; dfn[root] = low[root] = tmpdfn ; dfs(root) ; sumfz = countfz() ; memset(d , 0 , sizeof(d)) ; for(i = 0 ; i < cbridges ; i ++) // 统计各个边连通分量的度 { int ta = bridges[i][0] ; int tb = bridges[i][1] ; d[belong[ta]] ++ ; d[belong[tb]] ++ ; } int sumd1 = 0 ; for(i = 0 ; i < sumfz ; i ++) { if(d[i] == 1) sumd1 ++ ; } printf("%d\n" , (sumd1 + 1) / 2) ; // （度数为1的顶点个数 + 1）/ 2 即得答案 } int main() { while (scanf("%d%d" , &n , &m) != EOF) { clr() ; init() ; solve() ; } return 0 ; }