HDU 2686 Matrix

Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 904 Accepted Submission(s): 508

Problem Description

Yifenfei 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 yifenfei should to do is that choose a detour which frome 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 yifenfei 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 yifenfei can not pass the same area of the Matrix except the start and end. 

Input

The input contains multiple test cases.
Each case first line given the integer n (2<n<30) 
Than n lines,each line include n positive integers.(<100)

Output

For each test case output the maximal values yifenfei 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
   
   
   
   

Author

yifenfei

Source

ZJFC 2009-3 Programming Contest

Recommend

yifenfei

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <list>

#define INT_INF 0x3fffffff
#define LL_INF 0x3fffffffffffffff
#define EPS 1e-12
#define MOD 1000000007
#define PI 3.141592653579798
#define N 2000
#define E 100000

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;

struct Edge
{
    int st,en,cap,flow,cost,next;
} edge[E];
int head[N] , tot , now[N];
int source,sink;
int pre[N] , dis[N];
queue<int> q;
bool vs[N];

void add_edge(int st,int en,int cap,int cost)
{
    edge[tot].st=st;
    edge[tot].en=en;
    edge[tot].cap=cap;
    edge[tot].flow=0;
    edge[tot].cost=cost;
    edge[tot].next=head[st];
    head[st]=tot++;

    edge[tot].st=en;
    edge[tot].en=st;
    edge[tot].cap=0;
    edge[tot].flow=0;
    edge[tot].cost=-cost;
    edge[tot].next=head[en];
    head[en]=tot++;
}

bool SPFA()
{
    for(int i=0; i<N; i++)
        dis[i]=-1;
    memset(vs,0,sizeof(vs));
    memset(now,-1,sizeof(now));
    while(!q.empty()) q.pop();
    q.push(source); dis[source]=0; vs[source]=1;
    while(!q.empty())
    {
        int u=q.front(); q.pop(); vs[u]=0;
        for(int i=head[u],v; i!=-1; i=edge[i].next)
            if(edge[i].cap-edge[i].flow>0 && dis[v=edge[i].en]<dis[u]+edge[i].cost)
            {
                dis[v]=dis[u]+edge[i].cost;
                now[v]=i;
                if(!vs[v])
                {
                    vs[v]=1;
                    q.push(v);
                }
            }
    }
    if(dis[sink]!=-1) return true;
    else return false;
}

int MCMF()
{
    int cost=0;
    while(SPFA())
    {
        int flow=INT_INF;
        for(int u=sink; u!=source; u=edge[now[u]].st)
            if(flow>edge[now[u]].cap-edge[now[u]].flow)
                flow=edge[now[u]].cap-edge[now[u]].flow;
        for(int u=sink; u!=source; u=edge[now[u]].st)
        {
            edge[now[u]].flow+=flow;
            edge[now[u]^1].flow-=flow;
        }
        cost+=flow*dis[sink];
    }
    return cost;
}

void build(int n)
{
    memset(head,-1,sizeof(head));
    tot=0;
    source=N-2; sink=N-1;
    for(int i=0; i<n; i++)
        for(int j=0,val,p1,p2; j<n; j++)
        {
            p1=i*n+j;
            p2=p1+n*n;
            scanf("%d",&val);
            add_edge(p1,p2,1,val);
            if(p1==0 || p1==n*n-1) add_edge(p1,p2,INT_INF,0);
            if(i+1<n) add_edge(p2,(i+1)*n+j,1,0);
            if(j+1<n) add_edge(p2,i*n+j+1,1,0);
        }
    add_edge(source,0,2,0);
    add_edge(n*n-1+n*n,sink,2,0);
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        build(n);
        int ans=MCMF();
        printf("%d\n",ans);
    }
    return 0;
}