Matrix Again

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)

Problem Description

Starvae very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.

Every time starvae should to do is that choose a detour which from the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix starvae choose. But from the top to the bottom can
only choose right and down, from the bottom to the top can only choose left and up. And starvae can not pass the same area of the Matrix except the start and end..

Do you know why call this problem as “Matrix Again”? AS it is like the problem 2686 of HDU.

Input

The input contains multiple test cases.

Each case first line given the integer n (2<=n<=600)

Then n lines, each line include n positive integers. (<100)

Output

For each test case output the maximal values starvae can get.

Sample Input

2
10 3
5 10
3
10 3 3
2 5 3
6 7 10
5
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9

Sample Output

28
46
80

分析：这道题和hdu 2628（Matrix）一模一样，只不过数据变大了，我一开始认为要卡时间或卡内存，可不知怎么优化，于是抱着试试的心态把数组开大结果过了。

那大家可以看一下我的hdu 2628 有讲解。

代码示例:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define Min(a,b) a<b?a:b
#define inf 0xffffff
#define maxn 4000000+10
#define maxm 1000000+10
using namespace std;
typedef struct
{
    int from,to,next;
    int val,cost;
}node;
node E[maxn];
int head[maxm],dis[maxm],pre[maxm],pos[maxm],visit[maxm],cnt;
void init()
{
    memset(head,-1,sizeof(head));
    cnt=0;
}
void add(int from,int to,int val,int cost)
{
    E[cnt].from=from,E[cnt].to=to,E[cnt].val=val,E[cnt].cost=cost;
    E[cnt].next=head[from],head[from]=cnt++;
    E[cnt].from=to,E[cnt].to=from,E[cnt].val=0,E[cnt].cost=-cost;
    E[cnt].next=head[to],head[to]=cnt++;
}
bool spfa(int s,int t,int n)
{
    int val,cost,to;
    for(int i=0;i<=n;i++)
    {
        dis[i]=inf,visit[i]=0,pre[i]=-1;
    }
    dis[s]=0,visit[s]=1,pre[s]=s;
    queue<int>Q;
    Q.push(s);
    while(!Q.empty())
    {
        int k=Q.front();
        Q.pop();
        visit[k]=0;
        for(int i=head[k];i!=-1;i=E[i].next)
        {
            to=E[i].to,val=E[i].val,cost=E[i].cost;
            if(val>0&&dis[k]+cost<dis[to])
            {
                dis[to]=dis[k]+cost;
                pre[to]=k;
                pos[to]=i;
                if(visit[to])
                continue;
                visit[to]=1;
                Q.push(to);
            }
        }
    }
    if(pre[t]!=-1&&dis[t]<inf)
    return true;
    return false;
}
int MinCostFlow(int s,int t,int n)
{
    int netflow=0,costflow=0,min;
    while(spfa(s,t,n))
    {
        min=inf;
        for(int i=t;i!=s;i=pre[i])
        {
            min=Min(min,E[pos[i]].val);
        }
        netflow+=min;
        costflow+=min*dis[t];
        for(int i=t;i!=s;i=pre[i])
        {
            E[pos[i]].val-=min;
            E[pos[i]^1].val+=min;
        }
    }
    return -costflow;
}
int main()
{
    int n,s,t,x,sum;
    while(~scanf("%d",&n))
    {
        init();
        sum=0;
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        {
            scanf("%d",&x);
            if((i==0&&j==0)||(i==n-1&&j==n-1))
            {
                add(i*n+j,i*n+j+n*n,2,-x);
                sum+=x;
            }
            else
            {
                add(i*n+j,i*n+j+n*n,1,-x);
            }
            if(i+1<n)
            {
                add(i*n+j+n*n,(i+1)*n+j,1,0);
            }
            if(j+1<n)
            {
                add(i*n+j+n*n,i*n+j+1,1,0);
            }
        }
        s=2*n*n+1,t=2*n*n+2;
        add(s,0,2,0);
        add(2*n*n-1,t,2,0);
        printf("%d\n",MinCostFlow(s,t,t+10)-sum);
    }
    return 0;
}