codeforces 277E - Binary Tree on Plane （MinCostMaxFlow， 简单）

题意：
。。。
思路：
将每个点 u 拆成两个顶点 u-in, u-out， 但不连边。
从s向u-in连边， 容量为2， 限制u的子节点最大为2。
从u-out向t连边， 容量为1， 限制u只有一个父亲。
对 y[u]>y[v], 连一条边 （u， v）， cost 为距离。
然后跑 MinCostMaxFlow， 最后check一下是否满流。

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <map>#include <algorithm>#include <climits>#include <set>#include <queue>using namespace std;typedef long long LL;#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))const int N = 400;const double INF = 1e10;const double EPS = 1e-9;const int V = N * 2 + 2;const int MAXE = V * V;struct Edge {    int cap, nxt, to;    double cost;};class minCostMaxFlow {public:    Edge E[MAXE];    int edges, head[V], bak[V], bake[V];    double dis[V];    bool vis[V];    void init() {        edges = 0;        memset(head, -1, sizeof(head));    }    void add_edge(int u, int v, double w, int c) {        E[edges] = (Edge) {c, head[u], v, w};        E[edges + 1] = (Edge) {0, head[v], u, -w};        head[u] = edges;        head[v] = edges + 1;        edges += 2;    }    double sol(int source, int target) {        double answer = 0;        while (true) {            for (int i = 0; i <= target; ++i) {                vis[i] = 0, dis[i] = INF;            }            queue<int> q;            vis[source] = 1;            dis[source] = 0;            q.push(source);            memset(bak, -1, sizeof(bak));            memset(bake, -1, sizeof(bake));            while (!q.empty()) {                int u = q.front(); q.pop();                vis[u] = 0;                for (int i = head[u]; i != -1;) {                    Edge &e = E[i];                    int v = e.to;                    if ( e.cap > 0 && dis[u] + e.cost + EPS < dis[v] ) {                        dis[v] = dis[u] + e.cost;                        bak[v] = u;                        bake[v] = i;                        if ( !vis[v] ) {                            vis[v] = 1;                            q.push(v);                        }                    }                    i = e.nxt;                }            }            if ( bak[target] == -1 ) break;            answer += dis[target];            for (int i = target; i != source; i = bak[i]) {                E[bake[i]].cap -= 1;                E[bake[i]^1].cap += 1;            }        }        return answer;    }    bool check(int n, int root, int target) {        for (int i = 0; i < n; ++i)            if ( i != root ) {                int u = i + n;                for (int j = head[u]; j != -1;) {                    Edge &e = E[j];                    if ( e.to == target ) {                        if ( e.cap > 0 ) return 0;                    }                    j = e.nxt;                }            }        return 1;    }};int n, x[N], y[N];minCostMaxFlow sol;double dist(int u, int v) {    double dx = x[u] - x[v], dy = y[u] - y[v];    return sqrt(dx * dx + dy * dy);}int main() {#ifdef _LOCA_ENV_    freopen("input.in", "r", stdin);#endif // _LOCA_ENV    scanf("%d", &n);    for (int i = 0; i < n; ++i) {        scanf("%d%d", x + i, y + i);    }    int root_id = max_element(y, y + n) - y;    int source = n << 1, target = n << 1 | 1;    sol.init();    for (int i = 0; i < n; ++i) {        sol.add_edge(source, i, 0, 2);        sol.add_edge(i + n, target, 0, 1);    }    for (int i = 0; i < n; ++i)        for (int j = 0; j < n; ++j)            if ( y[i] > y[j] ) {                sol.add_edge(i, j + n, dist(i, j), 1);            }    double answer = sol.sol(source, target);    if ( !sol.check(n, root_id, target) ) {            printf("-1");            return 0;    }    printf("%.10f", answer);    return 0;}