Cable TV Network
The interconnection of the relays in a cable TV network is bi-directional. The network is connected if there is at least one interconnection path between each pair of relays present in the network. Otherwise the network is disconnected. An empty network or a network with a single relay is considered connected. The safety factor f of a network with n relays is:
1. n, if the net remains connected regardless the number of relays removed from the net.
2. The minimal number of relays that disconnect the network when removed.
For example, consider the nets from figure 1, where the circles mark the relays and the solid lines correspond to interconnection cables. The network (a) is connected regardless the number of relays that are removed and, according to rule (1), f=n=3. The network (b) is disconnected when 0 relays are removed, hence f=0 by rule (2). The network (c) is disconnected when the relays 1 and 2 or 1 and 3 are removed. The safety factor is 2.
Input
Write a program that reads several data sets from the standard input and computes the safety factor for the cable networks encoded by the data sets. Each data set starts with two integers: 0<=n<=50,the number of relays in the net, and m, the number of cables in the net. Follow m data pairs (u,v), u < v, where u and v are relay identifiers (integers in the range 0..n-1). The pair (u,v) designates the cable that interconnects the relays u and v. The pairs may occur in any order.Except the (u,v) pairs, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.
Output
For each data set, the program prints on the standard output, from the beginning of a line, the safety factor of the encoded net.
Sample Input
0 0
1 0
3 3 (0,1) (0,2) (1,2)
2 0
5 7 (0,1) (0,2) (1,3) (1,2) (1,4) (2,3) (3,4)
Sample Output
0
1
3
0
2
Hint
The first data set encodes an empty network, the second data set corresponds to a network with a single relay, and the following three data sets encode the nets shown in figure 1.
Source
题目类型：顶点连通度(最小割)
算法分析：使用独立轨和Menger定理将求解顶点连通度的问题转化成求解最小割的问题，对于任意两点a、b之间的最大独立轨数目的求解可以使用网络流。建图方式为先将点i拆点为i’和i”，然后构造<i’, i”>，容量为1。对于原图中的<u, v>，构造<u”, v’>和<v”, u’>，容量为INF。最后枚举源点s”和汇点e’，计算最大流并更新res。若res == INF，则表示顶点连通度为n，否则为res。注意枚举源点和汇点时两个点不能相邻！！！
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/************************************************* Author :supermaker Created Time :2016/4/9 13:00:09 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 4e4 + 66; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 200 + 66; struct Node { int v, nxt, c, f; }; Node edge[maxn*maxn]; int head[maxn], edgelen; int dep[maxn], cnt[maxn], cur[maxn]; void Init () { memset (head, -1, sizeof (head)); memset (edge, -1, sizeof (edge)); edgelen = 0; } void AddEdge (int u, int v, int w, int rw = 0) { edge[edgelen].v = v, edge[edgelen].c = w, edge[edgelen].f = 0; edge[edgelen].nxt = head[u], head[u] = edgelen++; edge[edgelen].v = u, edge[edgelen].c = rw, edge[edgelen].f = 0; edge[edgelen].nxt = head[v], head[v] = edgelen++; } void bfs (int s, int e) { memset (dep, -1, sizeof (dep)); memset (cnt, 0, sizeof (cnt)); dep[e] = 0; cnt[dep[e]]++; queue <int> qu; qu.push (e); while (!qu.empty ()) { int u = qu.front (); qu.pop (); for (int i = head[u]; i != -1; i = edge[i].nxt) { int v = edge[i].v; if (dep[v] != -1) continue; qu.push (v); dep[v] = dep[u] + 1; cnt[dep[v]]++; } } } int sta[maxn]; int MaxFlow (int s, int e, int n) { bfs (s, e); int res = 0, top = 0, u = s; memcpy (cur, head, sizeof (head)); while (dep[s] < n) { if (u == e) { int minval = INF, minval_pos = -1; for (int i = 0; i < top; i++) if (minval > edge[sta[i]].c - edge[sta[i]].f) { minval = edge[sta[i]].c - edge[sta[i]].f; minval_pos = i; } for (int i = 0; i < top; i++) { edge[sta[i]].f += minval; edge[sta[i]^1].f -= minval; } res += minval; top = minval_pos; u = edge[sta[top]^1].v; } else { bool is_have = false; for (int i = cur[u]; i != -1; i = edge[i].nxt) { int v = edge[i].v; if (edge[i].c - edge[i].f && dep[v] + 1 == dep[u]) { cur[u] = i; sta[top++] = cur[u]; u = v; is_have = true; break; } } if (is_have) continue; int minval = n; for (int i = head[u]; i != -1; i = edge[i].nxt) if (edge[i].c - edge[i].f && dep[edge[i].v] < minval) { minval = dep[edge[i].v]; cur[u] = i; } cnt[dep[u]]--; if (!cnt[dep[u]]) return res; dep[u] = minval + 1; cnt[dep[u]]++; if (u != s) u = edge[sta[--top]^1].v; } } return res; } int n, m; bool g[maxn][maxn]; void Build () { Init (); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { if (i == j) AddEdge (i, n + i, 1); else if (g[i][j]) AddEdge (n + i, j, INF); } } int main() { //CFF; //CPPFF; while (scanf ("%d%d", &n, &m) != EOF) { memset (g, false, sizeof (g)); for (int i = 1; i <= m; i++) { int u, v; scanf (" (%d,%d) ", &u, &v); u++, v++; g[u][v] = g[v][u] = true; } int res = INF; for (int i = 1; i <= n; i++) for (int j = i + 1; j <= n; j++) { if (!g[i][j]) { Build (); int tt = MaxFlow (n + i, j, 2 * n); res = min (res, tt); } } if (res >= INF) printf ("%d\n", n); else printf ("%d\n", res); } return 0; } |
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