HDU 3376 Matrix Again (EK)

Matrix Again

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 2357    Accepted Submission(s): 725

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

 

 

Author

Starvae

 

 

Source

HDOJ Monthly Contest – 2010.04.04

 

 

Recommend

lcy

 

 

 

用网络流写了下，拆点建边，费用为该点值的负数，源点和汇点容量2，其他边容量1；

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int INF=0x3f3f3f3f;
const int VM=800010;
const int EM=3000010;

struct Edge{
    int u,v,nxt;
    int flow,cap;
}edge[EM];

int n,cnt,head[VM],dis[VM],pre[VM];
int map[620][620],vis[VM];

void addedge(int cu,int cv,int cw,int cf){
    edge[cnt].u=cu;  edge[cnt].v=cv;  edge[cnt].cap=cw;
    edge[cnt].flow=cf;  edge[cnt].nxt=head[cu];  head[cu]=cnt++;
    edge[cnt].u=cv;  edge[cnt].v=cu;  edge[cnt].cap=0;
    edge[cnt].flow=-cf;  edge[cnt].nxt=head[cv];  head[cv]=cnt++;
}

int src,des;

int EK(){
    queue<int> q;
    while(!q.empty())
        q.pop();
    int ans=0;
    while(1){
        for(int i=0;i<=des;i++){
            dis[i]=INF;
            vis[i]=0;
        }
        dis[src]=0;
        q.push(src);
        pre[src]=-1;
        while(!q.empty()){
            int u=q.front();
            q.pop();
            vis[u]=0;
            for(int i=head[u];i!=-1;i=edge[i].nxt){
                int v=edge[i].v;
                if(edge[i].cap>0 && dis[v]>dis[u]+edge[i].flow){
                    dis[v]=dis[u]+edge[i].flow;
                    pre[v]=i;
                    if(!vis[v]){
                        vis[v]=1;
                        q.push(v);
                    }
                }
            }
        }
        if(dis[des]==INF)
            break;
        for(int u=des;pre[u]!=-1;u=edge[pre[u]].u){
            edge[pre[u]].cap--;
            edge[pre[u]^1].cap++;
        }
        ans+=dis[des];
    }
    return -ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n)){
        cnt=0;
        memset(head,-1,sizeof(head));
        src=0;  des=2*n*n-1;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&map[i][j]);
        addedge(src,n*n,2,0);
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++){
                if(i+1<n)
                    addedge(n*(n+i)+j,n*(i+1)+j,1,0);
                if(j+1<n)
                    addedge(n*(n+i)+j,n*i+(j+1),1,0);
                if((i==0 && j==0) || (i==n-1 && j==n-1))
                    continue;
                addedge(i*n+j,n*(n+i)+j,1,-map[i][j]);
            }
        addedge(n*n-1,2*n*n-1,2,0);
        printf("%d\n",EK()+map[0][0]+map[n-1][n-1]);
    }
    return 0;
}