POJ 1966 Cable TV Network（无向图的顶点连通度）

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POJ 1966 Cable TV Network

链接：http://poj.org/problem?id=1966

题意：有线电视网络中，中继器的连接是双向的。如果网络中任何两个中继器之间至少有一条路，则中继器网络称为是连通的，否则中继器网络是不连通的。一个空的网络、以及只有一个中继器的网络被认为是连通的。具有n 个中继器的网络的安全系数f 被定义成：
(1) f 为n，如果不管删除多少个中继器，剩下的网络仍然是连通的；
(2) f 为删除最少的顶点数，使得剩下的网络不连通。

现在给定一个有线电视网络，求 f 为多少。

思路：一张连通图，求最少删去多少个点，能够使得该图不再连通，这是无向图的顶点连通度问题。

无向图的点连通度：
1. 将每个点u拆为u', u''.顶点u'到u''连一条弧，容量为1。
2. 将图中的每条边(u, v)拆成<u'', v'>和<v'', u'>两条边，每条边的容量为INF。
3. 选一个源点A'', 枚举汇点B'. 求出最大流的最小值即为点连通度。
4. 所有具有流量1的弧(v', v'')对应的顶点v构成了一个割点集。

代码：

/*ID: wuqi9395@126.comPROG:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>using namespace std;#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define rep(i, a, n) for (int i = a; i < n; i++)#define per(i, a, n) for (int i = n - 1; i >= a; i--)#define eps 1e-6#define debug puts("===============")#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define fi first#define se second#define SZ(x) ((int)(x).size())#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)typedef long long ll;typedef unsigned long long ULL;const int maxn = 1000;const int maxm = 100000;int edge[maxm][2];struct node {    int v;    // vertex    int cap;    // capacity    int flow;   // current flow in this arc    int nxt;} e[maxm * 2];int g[maxn], cnt;int st, ed, n, m;void add(int u, int v, int c) {    e[++cnt].v = v;    e[cnt].cap = c;    e[cnt].flow = 0;    e[cnt].nxt = g[u];    g[u] = cnt;    e[++cnt].v = u;    e[cnt].cap = 0;    e[cnt].flow = 0;    e[cnt].nxt = g[v];    g[v] = cnt;}void init(int N) {    cnt = 1;    st = N;    for (int i = 0; i < n; i++) g[i] = 0;    int u, v;    for (int i = 0; i < N; i++) add(i, i + N, 1);    for (int i = 0; i < m; i++) {        int u = edge[i][0], v = edge[i][1];        add(v + N, u, INF);        add(u + N, v, INF);    }}int dist[maxn], numbs[maxn], q[maxn];void rev_bfs() {    int font = 0, rear = 1;    for (int i = 0; i <= n; i++) { //n为总点数        dist[i] = maxn;        numbs[i] = 0;    }    q[font] = ed;    dist[ed] = 0;    numbs[0] = 1;    while(font != rear) {        int u = q[font++];        for (int i = g[u]; i; i = e[i].nxt) {            if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;            dist[e[i].v] = dist[u] + 1;            ++numbs[dist[e[i].v]];            q[rear++] = e[i].v;        }    }}int maxflow() {    rev_bfs();    int u, totalflow = 0;    int curg[maxn], revpath[maxn];    for(int i = 0; i <= n; ++i) curg[i] = g[i];    u = st;    while(dist[st] < n) {        if(u == ed) {   // find an augmenting path            int augflow = INF;            for(int i = st; i != ed; i = e[curg[i]].v)                augflow = min(augflow, e[curg[i]].cap);            for(int i = st; i != ed; i = e[curg[i]].v) {                e[curg[i]].cap -= augflow;                e[curg[i] ^ 1].cap += augflow;                e[curg[i]].flow += augflow;                e[curg[i] ^ 1].flow -= augflow;            }            totalflow += augflow;            u = st;        }        int i;        for(i = curg[u]; i; i = e[i].nxt)            if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;        if(i) {   // find an admissible arc, then Advance            curg[u] = i;            revpath[e[i].v] = i ^ 1;            u = e[i].v;        } else {    // no admissible arc, then relabel this vertex            if(0 == (--numbs[dist[u]])) break;    // GAP cut, Important!            curg[u] = g[u];            int mindist = n;            for(int j = g[u]; j; j = e[j].nxt)                if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);            dist[u] = mindist + 1;            ++numbs[dist[u]];            if(u != st)                u = e[revpath[u]].v;    // Backtrack        }    }    return totalflow;}int main () {    int N;    while(~scanf("%d%d", &N, &m)) {        n = 2 * N + 4;        for (int i = 0; i < m; i++) scanf(" (%d,%d)", edge[i], edge[i] + 1);        int ans = INF;        for (int i = 1; i < N; i++) {            ed = i;            init(N);            ans = min(ans, maxflow());        }        if (ans == INF) ans = N;        printf("%d\n", ans);    }    return 0;}

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