UVALive 7037 (最大密度子图 网络流)

每个数都看成一个节点,每个逆序对之间的节点连边,于是只需要求出这个图的最大密度子图,可以用最小割模型解决.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const double INF = 1e8;
#define maxn 1511
#define maxm 2111111
#define eps 1e-10
#define type double

int n, m, x, N;
int s, t;
struct Edge
{
    int from, to,next;
    type cap,flow;
    void get(int u,int a,int b,type c,type d)
    {
        from = u; to = a; next = b; cap = c; flow = d;
    }
}edge[maxm];
int tol;
int head[maxn];
int gap[maxn],dep[maxn],pre[maxn],cur[maxn];
void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}

void add_edge(int u,int v,type w,type rw=0)
{ //cout << u << " " << v << " " << w << endl;
    edge[tol].get(u, v,head[u],w,0);head[u]=tol++;
    edge[tol].get(v, u,head[v],rw,0);head[v]=tol++;
}
type sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u=start;
    pre[u]=-1;
    gap[0]=N;
    type ans=0;
    while(dep[start]<N)
    {
        if(u==end)
        {
            type Min=INF;
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
                if(Min>edge[i].cap-edge[i].flow)
                   Min=edge[i].cap-edge[i].flow;
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
            {
                edge[i].flow+=Min;
                edge[i^1].flow-=Min;
            }
            u = start;
            ans+=Min;
            continue;
        }
        bool flag=false;
        int v;
        for(int i=cur[u];i !=-1;i=edge[i].next)
        {
            v=edge[i].to;
            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])
            {
                flag=true;
                cur[u]=pre[v]=i;
                break;
            }
        }
        if(flag)
        {
            u=v;
            continue;
        }
        int Min=N;
        for(int i=head[u];i!=-1;i=edge[i].next)
            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)
        {
            Min=dep[edge[i].to];
            cur[u]=i;
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u]=Min+1;
        gap[dep[u]]++;
        if(u!=start) u=edge[pre[u]^1].to;
    }
    return ans;
}

int degree[maxn];
int num[maxn];

type f (type g) { 
    init ();
    s = 0, t = n+1, N = n+2;
    double U = 1.0*m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j < i; j++) {
            if (num[j] > num[i]) {
                add_edge (i, j, 1);
                add_edge (j, i, 1);
            }
        }
    }
    for (int i = 1; i <= n; i++)
        add_edge (s, i, U);
    for (int i = 1; i <= n; i++)
        add_edge (i, t, U+2*g-1.0*degree[i]);
    double ans = sap (s, t, N); 
    ans = (U*n-ans)/2;
    return ans;
}

bool vis[maxn];

void dfs (int u) {
    vis[u] = 1;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if (vis[v])
            continue;
        if (edge[i].cap-edge[i].flow > 0 && edge[i].cap > 0)
            dfs (v);
    }
}

int main () {
    //freopen ("in.txt", "r", stdin);
    int tt, kase = 0;
    scanf ("%d", &tt);
    while (tt--) {
        scanf ("%d", &n);
        m = 0;
        memset (degree, 0, sizeof degree);
        for (int i = 1; i <= n; i++) {
            scanf ("%d", &num[i]);
            for (int j = 1; j < i; j++) {
                if (num[j] > num[i]) {
                    degree[i]++;
                    degree[j]++;
                    m++;
                }
            }
        }
        double l = 0, r = 10000;
        int kk = 50;
        while (kk--) {
            double mid = (l+r)/2;
            double h = f (mid);
            if (h <= 0)
                r = mid;
            else
                l = mid;
        }
        printf ("Case #%d: %.7f\n", ++kase, l);
    }
    return 0;
}