HDU 2686 MCMF

题意：两遍最长路，不能走重复点。和UVA 10806类似。

分析：拆点，u->u'，MCMF，求的是最大流的最小费用，那么cost取负。

注意的是源点，源点不用拆，那么走出来的最小费用，左上角的点，右下角的点走了两遍，输出除去即可。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2500+5;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, cap, flow, cost;
};

struct MCMF
{
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];         // 是否在队列中
    int d[maxn];           // Bellman-Ford
    int p[maxn];           // 上一条弧
    int a[maxn];           // 可改进量

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

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

    bool BellmanFord(int s, int t, int &flow, long long& cost)
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<n;i++)
            d[i] = INF;
        d[s] = 0;
        inq[s] = true;
        p[s] = 0;
        a[s] = INF;

        queue<int> Q;
        Q.push(s);
        while(!Q.empty())
        {
            int u = Q.front();
            Q.pop();
            inq[u] = false;
            for(int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
                {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if(!inq[e.to])
                    {
                        Q.push(e.to);
                        inq[e.to] = true;
                    }
                }
            }
        }
        if(d[t] == INF) return false; //s-t 不连通，失败退出
        flow += a[t];
        cost += (long long)d[t] * (long long)a[t];
        int u = t;
        while(u != s)
        {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }

    long long Mincost(int s, int t)
    {
        long long cost = 0;
        int flow = 0;
        while(BellmanFord(s, t, flow, cost)) {
            if(flow==2)
                break;
        };
        return cost;
    }
}sol;

int maps[maxn][maxn];
int aa[maxn*maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n)!=EOF) {

        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&maps[i][j]);

        int s = 0,t = n*n-1;
        sol.init(n*n*2);

        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                int id = i*n + j;
                if(id!=s&&id!=t)
                    sol.AddEdge(id,id+n*n,1,-maps[i][j]);
                if(id==s) {
                    sol.AddEdge(id,id+1,1,0);
                    sol.AddEdge(id,id+n,1,0);
                }
                else {
                    if(i<n-1) sol.AddEdge(id+n*n,id+n,1,0);
                    if(j<n-1) sol.AddEdge(id+n*n,id+1,1,0);
                }
            }
        }

        printf("%d\n",-sol.Mincost(s,t)+maps[0][0]+maps[n-1][n-1]);



    }
    return 0;
}