HDU3376 最小费用最大流 模板2

Matrix Again

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 4255    Accepted Submission(s): 1233

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

 代码：

//和HDU2686一样，只是数据变大了，上一个模板会超内存,这个板不会。
/***********************最小费用最大流模板2*************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=**+;
const int maxm=*maxn;//！边数要够
const int inf=0x7fffffff;
struct Edge
{
    int to,next,cap,flow,cost;
}edges[maxm];
int head[maxn],tol,pre[maxn],dis[maxn];
bool vis[maxn];
int N;
void init(int n)
{
    N=n;
    tol=;
    memset(head,-,sizeof(head));
}
void AddEdge(int u,int v,int cap,int cost)
{
    edges[tol].to=v;
    edges[tol].cap=cap;
    edges[tol].cost=cost;
    edges[tol].flow=;
    edges[tol].next=head[u];
    head[u]=tol++;
    edges[tol].to=u;
    edges[tol].cap=;
    edges[tol].cost=-cost;
    edges[tol].flow=;
    edges[tol].next=head[v];
    head[v]=tol++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    for(int i=;i<=N;i++){
        dis[i]=inf;
        vis[i]=;
        pre[i]=-;
    }
    dis[s]=;
    vis[s]=;
    q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();
        vis[u]=;
        for(int i=head[u];i!=-;i=edges[i].next){
            int v=edges[i].to;
            if(edges[i].cap>edges[i].flow&&dis[v]>dis[u]+edges[i].cost){
                dis[v]=dis[u]+edges[i].cost;
                pre[v]=i;
                if(!vis[v]) {vis[v]=;q.push(v);}
            }
        }
    }
    if(pre[t]==-) return ;
    return ;
}
int MinCostFlow(int s,int t)
{
    int flow=,cost=;
    while(spfa(s,t)){
        int Min=inf;
        for(int i=pre[t];i!=-;i=pre[edges[i^].to])
            Min=min(Min,edges[i].cap-edges[i].flow);
        for(int i=pre[t];i!=-;i=pre[edges[i^].to]){
            edges[i].flow+=Min;
            edges[i^].flow-=Min;
            cost+=edges[i].cost*Min;
        }
        flow+=Min;
    }
    return cost;//返回最小费用，flow存最大流
}
/*********************************************************************/
int main()
{
    int n,mp[][];
    while(scanf("%d",&n)==){
        for(int i=;i<=n;i++)
            for(int j=;j<=n;j++) scanf("%d",&mp[i][j]);
        int s=,t=n*n*+;
        init(n*n*+);
        for(int i=;i<=n;i++){
            for(int j=;j<=n;j++){
                int id=(i-)*n+j;
                if(id==){
                    AddEdge(id,id+n*n,,-mp[i][j]);
                    AddEdge(s,id,,);
                }
                else if(id==n*n){
                    AddEdge(id,id+n*n,,-mp[i][j]);
                    AddEdge(id+n*n,t,,);
                }
                else AddEdge(id,id+n*n,,-mp[i][j]);
                if(i<n) AddEdge(id+n*n,id+n,,);
                if(j<n) AddEdge(id+n*n,id+,,);
            }
        }
        int ans=-(MinCostFlow(s,t)+mp[][]+mp[n][n]);
        printf("%d\n",ans);
    }
    return ;
}