欢迎来到 xht37 的博客

UVA1660 电视网络 Cable TV Network

题目地址:UVA1660 电视网络 Cable TV Network

枚举两个不直接连通的点 \(S\)\(T\) ,求在剩余的 \(n-2\) 个节点中最少去掉多少个可以使 \(S\)\(T\) 不连通,在每次枚举的结构中取 \(min\) 就是本题的答案。

点边转化

把原来无向图中的每个点 \(x\) ,拆成入点 \(x\) 和出点 \(x'\) 。在无向图中删去一个点⇔在网络中断开 \((x,x')\) 。对 \(\forall x \neq S,x \neq T\) 连有向边 \((x,x')\) ,容量为 \(1\) 。对原无向图的每条边 \((x,y)\) ,连有向边 \((x',y),(y',x)\) ,容量为 \(+ \infty\) (防止割断)。

求最小割即可。

#include <bits/stdc++.h>
using namespace std;
const int N = 56, M = 2e4 + 6, inf = 0x3f3f3f3f;
int n, m, s, t;
int a[N*N], b[N*N], d[N<<1];
int Head[N<<1], Edge[M], Leng[M], Next[M], tot;

inline void add(int x, int y, int z) {
    Edge[++tot] = y;
    Leng[tot] = z;
    Next[tot] = Head[x];
    Head[x] = tot;
    Edge[++tot] = x;
    Leng[tot] = 0;
    Next[tot] = Head[y];
    Head[y] = tot;
}

inline bool bfs() {
    memset(d, 0, sizeof(d));
    queue<int> q;
    q.push(s);
    d[s] = 1;
    while (q.size()) {
        int x = q.front();
        q.pop();
        for (int i = Head[x]; i; i = Next[i]) {
            int y = Edge[i], z = Leng[i];
            if (z && !d[y]) {
                q.push(y);
                d[y] = d[x] + 1;
                if (y == t) return 1;
            }
        }
    }
    return 0;
}

inline int dinic(int x, int f) {
    if (x == t) return f;
    int rest = f;
    for (int i = Head[x]; i && rest; i = Next[i]) {
        int y = Edge[i], z = Leng[i];
        if (z && d[y] == d[x] + 1) {
            int k = dinic(y, min(rest, z));
            if (!k) d[y] = 0;
            Leng[i] -= k;
            Leng[i^1] += k;
            rest -= k;
        }
    }
    return f - rest;
}

inline void Cable_TV_Network() {
    for (int i = 0; i < m; i++) {
        char str[20];
        scanf("%s", str);
        a[i] = b[i] = 0;
        int j;
        for (j = 1; str[j] != ','; j++) a[i] = a[i] * 10 + str[j] - '0';
        for (j++; str[j] != ')'; j++) b[i] = b[i] * 10 + str[j] - '0';
    }
    int ans = inf;
    for (s = 0; s < n; s++)
        for (t = 0; t < n; t++)
            if (s != t) {
                memset(Head, 0, sizeof(Head));
                tot = 1;
                int maxf = 0;
                for (int i = 0; i < n; i++)
                    if (i == s || i == t) add(i, i + n, inf);
                    else add(i, i + n, 1);
                for (int i = 0; i < m; i++) {
                    add(a[i] + n, b[i], inf);
                    add(b[i] + n, a[i], inf);
                }
                while (bfs()) {
                    int num;
                    while ((num = dinic(s, inf))) maxf += num;
                }
                ans = min(ans, maxf);
            }
    if (n <= 1 || ans == inf) ans = n;
    cout << ans << endl;
}

int main() {
    while (cin >> n >> m) Cable_TV_Network();
    return 0;
}
posted @ 2019-03-12 01:15 xht37 阅读(...) 评论(...) 编辑 收藏