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2019-2020 ACM-ICPC Brazil Subregional Programming Contest G Getting Confidence

Getting Confidence

费用流

经典费用流模型,先建立 \(n * n\) 个点,来模拟一行,一行中的连接边流量为 \(1\),费用为 \(0\),源点连接每一行的第一个点

对上述生成的点中,同一列连接到一个新的点,流量为 \(1\),费用为 \(-a[i][j]\),然后该点连接一个流量为 \(1\),费用为 \(0\) 的点到汇点

这样构造之后,相当于跑一个最大费用最大流,可以保证如果选取了一行中的一个点,列也就只能选一个

因为是乘法,数字特别大,因此可以考虑变成 \(log\) 之后换成加法来跑

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;
const int maxm = 2e5 + 10;
const double inf = 1e17;
const double eps = 1e-8;
int n, m, s, t, tp = 1;
int head[maxn], nex[maxm], to[maxm], cur[maxn], vis[maxn];
double dis[maxn], cost[maxm];
ll val[maxm];

inline int dcmp(double x)
{
    return (x > eps) - (x < -eps);
}

void add(int u, int v, ll f, double c)
{
    tp++;
    nex[tp] = head[u];
    head[u] = tp;
    to[tp] = v;
    cost[tp] = c;
    val[tp] = f;
}

bool spfa()
{
    for(int i=0; i<=t; i++) cur[i] = head[i];
    for(int i=0; i<=t; i++) dis[i] = inf;
    dis[s] = 0;
    queue<int>q;
    q.push(s);
    while(q.size())
    {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i=head[u]; i; i=nex[i])
        {
            int v = to[i];
            if(val[i] > 0 && dis[u] + cost[i] < dis[v])
            {
                dis[v] = dis[u] + cost[i];
                if(vis[v] == 0)
                {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return dcmp(inf - dis[t]);
}

double c_ans = 0;
ll dfs(int now, ll flow)
{
    if(now == t) return flow;
    vis[now] = 1;
    ll ans = 0;
    for(int i=cur[now]; i && ans < flow; i=nex[i])
    {
        cur[now] = i;
        int v = to[i];
        if(vis[v] == 0 && val[i] > 0 && dis[v] == dis[now] + cost[i])
        {
            ll x = dfs(v, min(val[i], flow - ans));
            c_ans += x * cost[i];
            val[i] -= x;
            val[i ^ 1] += x;
            ans += x;
        }
    }
    vis[now] = 0;
    return ans;
}

ll mcmf()
{
    ll ans = 0;
    while(spfa())
    {
        ll x = dfs(s, inf);
        ans += x;
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    s = n * n + n + 1;
    t = s + 1;
    int tx = 0;
    for(int i=0; i<n; i++)
    {
        int now = ++tx;
        add(s, now, 1, 0);
        add(now, s, 0, 0);
        for(int j=1; j<n; j++)
        {
            ++tx;
            add(now, tx, 1, 0);
            add(tx, now, 0, 0);
            now = tx;
        }
    }
    for(int i=1; i<=n*n; i++)
    {
        int g = (i - 1) % n + 1 + n * n;
        double x;
        cin >> x;
        x = log10(x);
        add(i, g, 1, -x);
        add(g, i, 0, x);
    }
    for(int i=1; i<=n; i++)
    {
        int g = i + n * n;
        add(g, t, 1, 0);
        add(t, g, 0, 0);
    }
    ll ax = mcmf();
    vector<int>ans(n);
    for(int i=1; i<=n*n; i++)
    {
        for(int j=head[i]; j; j=nex[j])
        {
            int nex = to[j];
            if(nex > n * n && nex != s && val[j] == 0)
                ans[(i - 1) % n] = (i - 1) / n + 1;
        }
    }
    for(int i=0; i<ans.size(); i++)
    {
        if(i) cout << " ";
        cout << ans[i];
    }
    cout << endl;
    return 0;
}
posted @ 2022-09-23 14:13  dgsvygd  阅读(103)  评论(0)    收藏  举报