P1345 [USACO5.4]奶牛的电信Telecowmunication

传送门

思路:给出一个无向图,问两点间至少需要破坏多少点,就能使得两点不连通。

思路:显然要用最大流解决,怎么建图呢,拆点,我们可以把一个点拆成

两个点,一个用来接受流量,一个用来发送流量,两点连边,流量设置为1,

然后再把边的流量也设置为1,注意是无向图,相反边也要建,这样跑最大流,

就能得到最少需要拆分的点的个数。

拆点不要大家不要怕,以为很高大上,操作起来和离散化一样其实是简单的,2x设置为接受点,2x+1设置为发送点即可

AC代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100005;
const int inf = 0x3f3f3f3f;
struct edge {
    int f, t, nxt;
    ll flow;
}e[maxn*2];
int hd[maxn], tot = 1;
void add(int f, int t, ll flow) {
    e[++tot] = { f,t,hd[f],flow };
    hd[f] = tot;
}
int n, m, c1,c2;
int s, d;
int dep[maxn], cur[maxn];
bool bfs() {//找增广路
    memset(dep, 0, sizeof(dep));
    dep[s] = 1;
    queue<int>Q;
    Q.push(s);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (int i = hd[u]; i; i = e[i].nxt) {
            int v = e[i].t, flow = e[i].flow;
            if (flow > 0 && !dep[v]) {
                dep[v] = dep[u] + 1;
                if (v == d)return true;
                Q.push(v);
            }
        }
    }
    return false;
}
ll dfs(int u, ll flow) {
    if (u == d)return flow;
    ll last = flow;
    for (int i = cur[u]; i; i = e[i].nxt) {
        cur[u] = i;//当前弧优化
        int v = e[i].t; ll flow = e[i].flow;
        if (flow > 0 && dep[v] == dep[u] + 1) {
            ll tmp = dfs(v, min(last, flow));
            last -= tmp;
            e[i].flow -= tmp;
            e[i ^ 1].flow += tmp;
            if (last == 0)break;//流量没了直接退出循环,与当前弧优化对应
        }
    }
    if (last == flow)dep[u] = 0;//从当前点一点流量没流到终点(未找到增广路),炸点优化
    return flow - last;//返回剩余流量
}
ll dinic() {
    ll maxflow = 0;
    while (bfs()) {
        memcpy(cur, hd, sizeof(hd));
        maxflow += dfs(s, inf);
    }
    return maxflow;
}
int main() {
    //freopen("test.txt", "r", stdin);
    scanf("%d%d%d%d", &n, &m, &c1, &c2);
    for (int i = 1; i <= n; i++) {
        add(i * 2, i * 2 + 1, 1);
        add(i * 2 + 1, i * 2, 0);//先拆点
    }
    for (int i = 1; i <= m; i++) {
        int a, b; scanf("%d%d", &a, &b);
        add(a*2+1, b*2, 1);
        add(b*2, a*2+1, 0);
        add(b*2+1, a*2, 1);//无向图,反向边也需要
        add(a*2, b*2+1, 0);
    }
    s = c1*2+1, d = c2*2;//起点是c1的发送点,终点是c2的接受点
    ll ans = dinic();
    printf("%lld\n", ans);
    return 0;
}

标签：last,int,USACO5.4,flow,dep,Telecowmunication,P1345,ll,hd
来源： https://www.cnblogs.com/MYMYACMer/p/14617867.html