2014-2015 ACM-ICPC, Asia Xian Regional Contest C – The Problem Needs 3D Arrays(最大密度子图)

题目大意：给你N个数，要求你找出k个数，使得xn/k的最大值。(xn代表的是k个数组成的逆序对的数量)

解题思路：将每个数当成一个点，如果两个数组成逆序对的关系，就连边，这样图就构成了
详解请看刘伯涛:最小割模型在信息学竞赛中的应用
这题的精度要求较高，要小心
这题还有一个技巧，就是二分确定S集，然后找出S级中的所有的边和点，然后除一下就可以了

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define N 1110
#define INF 0x3f3f3f3f
const double Max = 0x3fffffff;
const double esp = 1e-10;

struct Edge {
    int from, to;
    double cap, flow;
    Edge() {}
    Edge(int from, int to, double cap, double flow): from(from), to(to), cap(cap), flow(flow) {}
};

struct ISAP {
    int p[N], num[N], cur[N], d[N];
    int t, s, n, m;
    bool vis[N];

    vector<int> G[N];
    vector<Edge> edges;

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++) {
            G[i].clear();
            d[i] = INF;
        }
        edges.clear();
    }

    void AddEdge(int from, int to, double cap) {
        edges.push_back(Edge(from, to, cap, 0));
        edges.push_back(Edge(to, from, 0, 0));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool BFS() {
        memset(vis, 0, sizeof(vis));

        queue<int> Q;
        d[t] = 0;
        vis[t] = 1;
        Q.push(t);

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();

            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i] ^ 1];
                if (!vis[e.from] && e.cap - e.flow > esp) {
                    vis[e.from] = true;
                    d[e.from] = d[u] + 1;
                    Q.push(e.from);
                }
            }
        }
        return vis[s];
    }

    double Augment() {
        int u = t;
        double flow = Max;
        while (u != s) {
            Edge &e = edges[p[u]];
            flow = min(flow, e.cap - e.flow);
            u = edges[p[u]].from;
        }

        u = t;
        while (u != s) {
            edges[p[u]].flow += flow;
            edges[p[u] ^ 1].flow -= flow;
            u = edges[p[u]].from;
        }
        return flow;
    }

    double Maxflow(int s, int t) {
        this->s = s; this->t = t;
        double flow = 0;
        BFS();
        if (d[s] > n)
            return 0;

        memset(num, 0, sizeof(num));
        memset(cur, 0, sizeof(cur));
        for (int i = 0; i < n; i++)
            if (d[i] < INF)
                num[d[i]]++;
        int u = s;

        while (d[s] <= n) {
            if (u == t) {
                flow += Augment();
                u = s;
            }
            bool ok = false;
            for (int i = cur[u]; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap - e.flow > esp && d[u] == d[e.to] + 1) {
                    ok = true;
                    p[e.to] = G[u][i]; 
                    cur[u] = i;
                    u = e.to;
                    break;
                }
            }

            if (!ok) {
                int Min = n;
                for (int i = 0; i < G[u].size(); i++) {
                    Edge &e = edges[G[u][i]];
                    if (e.cap - e.flow > esp)
                        Min = min(Min, d[e.to]);
                }
                if (--num[d[u]] == 0)
                    break;
                num[d[u] = Min + 1]++;
                cur[u] = 0;
                if (u != s)
                    u = edges[p[u]].from;
            }
        }
        return flow;
    }
};

ISAP isap;
#define maxn 110
#define maxm 20010
int n, m, source, sink, Sum;
int x[maxm], y[maxm], in[maxn], num[maxn];
bool vis[maxn];

void init() {
    scanf("%d", &n);
    m = 0;
    memset(in, 0, sizeof(in));
    for (int i = 1; i <= n; i++)
        scanf("%d", &num[i]);

    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            if (num[i] > num[j]) {
                in[i]++; in[j]++;
                x[m] = i; y[m] = j;
                m++;
            }
}

void build(double mid) {
    isap.init(sink);
    for (int i = 1; i <= n; i++) {
        isap.AddEdge(source, i, m * 1.0);
        isap.AddEdge(i, sink, m + 2 * mid - in[i]);
    }

    for (int i = 0; i < m; i++) {
        isap.AddEdge(x[i], y[i], 1.0);
        isap.AddEdge(y[i], x[i], 1.0);
    }
}

void dfs(int u) {
    vis[u] = true;
    for (int i = 0; i < isap.G[u].size(); i++) {
        Edge &e = isap.edges[isap.G[u][i]];
        if (!vis[e.to] && e.cap - e.flow > esp) 
            dfs(e.to);
    }
}

int cas = 1; 
void solve() {
    if (m == 0) {
        printf("Case #%d: %.10f\n", cas++, 0.0);
        return ;
    }
    source = 0, sink = n + 1;

    double l = 0, r = m, mid, ans;
    while (r - l >= 1.0 / n / n){
        mid = (r + l) / 2;
        build(mid);
        ans = isap.Maxflow(source, sink);
        if ((1.0 * n * m - ans) / 2 > esp) l = mid;
        else r = mid;   
    }
    build(l);
    isap.Maxflow(source, sink);
    memset(vis, 0, sizeof(vis));
    dfs(source);

    int cnt1 = 0, cnt2 = 0;
    for (int i = 1; i <= n; i++)
        if (vis[i]) cnt2++;

    for (int i = 0; i < m; i++)
        if (vis[x[i]] && vis[y[i]]) cnt1++;
    printf("Case #%d: %.10f\n", cas++, cnt1 * 1.0 / cnt2);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}