hdu-3376-Matrix Again(最小费用最大流)

题意：

给一个矩形，从左上角走到右下角，并返回左上角（一个单元格只能走一次，左上角和右下角两个点除外）

并且从左上到右下只能往右和下两个方向。从右下返回左上只能走上和左两个方向！

分析：

拆点，最小费用最大流

。。

额。。。刘汝佳训练指南的最小费用最大流模板超时了。。。。。。。。。。。。。。。。。。

可能是因为边太少，点太多的缘故吧！还是数组实现的邻接表可靠啊！！！

// File Name: 3376.cpp
// Author: Zlbing
// Created Time: 2013年08月15日 星期四 13时24分37秒
 
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
/*
const int MAXN=720010;
 
struct Edge{
    int from,to,cap,flow,cost;
};
struct MCMF{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int> G[MAXN];
    int inq[MAXN];
    int d[MAXN];
    int p[MAXN];
    int a[MAXN];
    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){
        Edge e;
        e.from=from,e.to=to,e.cap=cap,e.flow=0,e.cost=cost;
        edges.push_back(e);
        //edges.push_back((Edge){from,to,cap,0,cost});
        e.from=to,e.to=from,e.cap=0,e.flow=0,e.cost=-cost;
        //edges.push_back((Edge){to,from,0,0,-cost});
        edges.push_back(e);
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BellmanFord(int s,int t,int& flow,int& cost){
        for(int i=0;i<=n;i++)d[i]=INF;
            CL(inq,0);
        d[s]=0;inq[s]=1;p[s]=0;a[s]=INF;
 
        queue<int>Q;
        Q.push(s);
        while(!Q.empty()){
            int u=Q.front();Q.pop();
            inq[u]=0;
            for(int i=0;i<(int)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]=1;
                    }
                }
            }
        }
        if(d[t]==INF)return false;
        flow+=a[t];
        cost+=d[t]*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;
    }
    int Mincost(int s,int t){
        int flow=0,cost=0;
        while(BellmanFord(s,t,flow,cost));
        return cost;
    }
};
MCMF solver;
*/
int sumFlow;
const int MAXN = ;
const int MAXM = ;
struct Edge
{
    int u, v, cap, cost;
    int next;
}edge[MAXM<<];
int NE;
int head[MAXN], dist[MAXN], pp[MAXN];
bool vis[MAXN];
void init()
{
    NE = ;
    memset(head, -, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
    edge[NE].u = u; edge[NE].v = v; edge[NE].cap = cap; edge[NE].cost = cost;
    edge[NE].next = head[u]; head[u] = NE++;
    edge[NE].u = v; edge[NE].v = u; edge[NE].cap = ; edge[NE].cost = -cost;
    edge[NE].next = head[v]; head[v] = NE++;
}
bool SPFA(int s, int t, int n)
{
    int i, u, v;
    queue <int> qu;
    memset(vis,false,sizeof(vis));
    memset(pp,-,sizeof(pp));
    for(i = ; i <= n; i++) dist[i] = INF;
    vis[s] = true; dist[s] = ;
    qu.push(s);
    while(!qu.empty())
    {
        u = qu.front(); qu.pop(); vis[u] = false;
        for(i = head[u]; i != -; i = edge[i].next)
        {
            v = edge[i].v;
            if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)
            {
                dist[v] = dist[u] + edge[i].cost;
                pp[v] = i;
                if(!vis[v])
                {
                    qu.push(v);
                    vis[v] = true;
                }
            }
        }
    }
    if(dist[t] == INF) return false;
    return true;
}
int MCMF(int s, int t, int n) // minCostMaxFlow
{
    int flow = ; // 总流量
    int i, minflow, mincost;
    mincost = ;
    while(SPFA(s, t, n))
    {
        minflow = INF + ;
        for(i = pp[t]; i != -; i = pp[edge[i].u])
            if(edge[i].cap < minflow)
                minflow = edge[i].cap;
        flow += minflow;
        for(i = pp[t]; i != -; i = pp[edge[i].u])
        {
            edge[i].cap -= minflow;
            edge[i^].cap += minflow;
        }
        mincost += dist[t] * minflow;
    }
    sumFlow = flow; // 题目需要流量，用于判断
    return mincost;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        //solver.init(n*n*2+5);
        init();
        int a;
        int s=;
        int t=((n-)*n+n-)*+;
        REP(i,,n-)
            REP(j,,n-){
                scanf("%d",&a);
                //solver.AddEdge((i*n+j)*2,(i*n+j)*2+1,1,-a);
                addedge((i*n+j)*,(i*n+j)*+,,-a);
                if(i!=n-)
                {
                    //solver.AddEdge((i*n+j)*2+1,((i+1)*n+j)*2,1,0);
                    addedge((i*n+j)*+,((i+)*n+j)*,,);
                }
                if(j!=n-)
                {
                    //solver.AddEdge((i*n+j)*2+1,(i*n+j+1)*2,1,0);
                    addedge((i*n+j)*+,(i*n+j+)*,,);
                }
            }
        addedge(s,s+,,);
        addedge(t-,t,,);
        int ans=-MCMF(s,t,t);
        printf("%d\n",ans);
    }
    return ;
}