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CF 277E Binary Tree on Plane

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题目描述

http://codeforces.com/problemset/problem/277/E

Solution

注意到河道题目中二叉树的限制为至多有两个儿子

另外注意到边权只跟父亲和儿子有关

那么我们考虑将所有点拆成两个点，第一个点表示作为父亲，第二个点表示作为儿子

最小费用最大流

建图：

$s$ 连 $i$，容量为 $2$，费用为 $0$

如果 $y_i>y_j$，$i$ 连 $j’$，容量为 $1$，费用为 两点的欧几里得距离

$i’$ 连 $t$，容量为 $1$，费用为 $0$

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#include <iostream>
#include <queue>
#include <iomanip>
#include <algorithm>
#include <cmath>
#define maxn 1010
#define INF 1000000000
using namespace std;

int n;

pair<int, int> a[maxn];

inline double F(int x) { return x * x; } 

inline double D(const pair<int, int> &u, const pair<int, int> &v) {
    return sqrt(F(u.first - v.first) + F(u.second - v.second));
}

struct Edge {
    int to, next, w;
    double fi; 
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w, double fi) {
    e[c1].to = v; e[c1].w = w; e[c1].fi = fi;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w, double fi) {
    add_edge(u, v, w, fi); add_edge(v, u, 0, -fi);
}

int la[maxn], pre[maxn], fl[maxn], s, t; double dis[maxn]; bool vis[maxn];
bool spfa() {
    fill(dis, dis + maxn, INF); dis[s] = 0;
    queue<int> Q; Q.push(s); pre[t] = -1; fl[s] = INF; vis[s] = 1;
    while (!Q.empty()) {
        int u = Q.front(); Q.pop(); vis[u] = 0; 
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w; double fi = e[i].fi;
            if (w > 0 && dis[v] > dis[u] + fi) {
                dis[v] = dis[u] + fi; la[v] = i;
                fl[v] = min(fl[u], w); pre[v] = u;
                if (vis[v]) continue; vis[v] = 1;
                Q.push(v);
            }
        }
    }
    return ~pre[t];
}

int mf; double mc;
void mcmf() {
    while (spfa()) {
        int now = t;
        mf += fl[t];
        mc += fl[t] * dis[t];
        while (now != s) {
            e[la[now]].w -= fl[t];
            e[la[now] ^ 1].w += fl[t];
            now = pre[now];
        }
    }
} 


int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i].first >> a[i].second;
    s = 0; t = 2 * n + 1;
    for (int i = 1; i <= n; ++i) {
        Add_edge(s, i, 2, 0);
        Add_edge(i + n, t, 1, 0);
        for (int j = 1; j <= n; ++j) 
            if (a[i].second > a[j].second)
                Add_edge(i, j + n, 1, D(a[i], a[j]));
    } mcmf(); 
    if (mf != n - 1) cout << "-1";
    else cout << fixed << setprecision(7) << mc << "\n"; 
    return 0;
}