POJ 1966 Cable TV Network (算竞进阶习题)

拆点+网络流

拆点建图应该是很常见的套路了。。一张无向图不联通，那么肯定有两个点不联通，但是我们不知道这两个点是什么。

所以我们枚举所有点，并把每个点拆成入点和出点，每次把枚举的两个点的入点作为s和t（这样方便，当然也可以把第一个点的出点当成s，第二个点的入点当成t，但其实我们把s和t的入点和出点之间的边容量设为INF之后就没有影响了）

每条原图的边连接着u的出点和v的入点，v的出点和u的入点，容量设为INF，保证不给割，其他点的入点和出点之间的容量当然是1。这样我们的割就一定会割在容量为1的边上，每割一条，就相当于去掉了一个点。暴力跑最大流早最小值就好了

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 1000;
int n, m, cnt, head[N<<2], x[N<<2], y[N<<2], s, t, depth[N<<2], ans;
struct Edge { int v, next, f; } edge[N<<5];

void addEdge(int a, int b, int f){
    edge[cnt].v = b, edge[cnt].f = f, edge[cnt].next = head[a], head[a] = cnt ++;
    edge[cnt].v = a, edge[cnt].f = 0, edge[cnt].next = head[b], head[b] = cnt ++;
}

bool bfs(){
    full(depth, 0);
    queue<int> q;
    depth[s] = 1, q.push(s);
    while(!q.empty()){
        int cur = q.front(); q.pop();
        //cout << cur << endl;
        for(int i = head[cur]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(!depth[u] && edge[i].f > 0){
                depth[u] = depth[cur] + 1;
                q.push(u);
            }
        }
    }
    return depth[t] != 0;
}

int dfs(int cur, int a){
    if(cur == t) return a;
    int flow = 0;
    for(int i = head[cur]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(depth[u] == depth[cur] + 1 && edge[i].f > 0){
            int k = dfs(u, min(a, edge[i].f));
            a -= k, flow += k, edge[i].f -= k, edge[i^1].f += k;
        }
        if(!a) break;
    }
    if(a) depth[cur] = -1;
    return flow;
}

int dinic(){
    int ret = 0;
    while(bfs()) ret += dfs(s, INF);
    return ret;
}

void solve(){
    full(head, -1), cnt = 0;
    for(int i = 0; i < n; i ++){
        if(i == s || i == t) addEdge(i, i + n, INF);
        else addEdge(i, i + n, 1);
    }
    for(int i = 0; i < m; i ++){
        addEdge(x[i] + n, y[i], INF), addEdge(y[i] + n, x[i], INF);
    }
    ans = min(ans, dinic());
}

int main(){

    while(scanf("%d%d", &n, &m) != EOF){
        for(int i = 0; i < m; i ++){
            int j; char str[100]; scanf("%s", str);
            for(x[i] = 0, j = 1; str[j] != ','; j ++) x[i] = x[i] * 10 + (str[j] - '0');
            for(y[i] = 0, j ++; str[j] != ')'; j ++) y[i] = y[i] * 10 + (str[j] - '0');
        }
        ans = INF;
        for(int i = 0; i < n; i ++){
            for(int j = i + 1; j < n; j ++){
                s = i, t = j, solve();
            }
        }
        if(n <= 1 || ans == INF) ans = n;
        printf("%d\n", ans);
    }
    return 0;
}