HDU 2686 Matrix(最大费用流)

Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1890    Accepted Submission(s): 1005

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

 题意：给一个n*n的距阵。每一个点都有一个值。问从(0,0)到(n-1, n-1)点（仅仅能从左到右 或 从上到下）再回到(0,0)点（仅仅能从右到左 或 下到上）经过的点的值总和最大是多少？每一个点仅仅能走一次。

解题：事实上就是找两条从(0,0)到(n-1,n-1)总和最大的路.拆点法。每一个边容量为1。费用：对于点的本身，边权为点权，非点的边权值为0。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN = 10010;
const int MAXM = 1001000;
const int INF = 1<<30;
struct EDG{
    int to,next,cap,flow;
    int cost;  //单位价格
}edg[MAXM];
int head[MAXN],eid;
int pre[MAXN], cost[MAXN] ; //点0~(n-1)

void init(){
    eid=0;
    memset(head,-1,sizeof(head));
}
void addEdg(int u,int v,int cap,int cst){
    edg[eid].to=v; edg[eid].next=head[u]; edg[eid].cost = cst;
    edg[eid].cap=cap; edg[eid].flow=0; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v]; edg[eid].cost = -cst;
    edg[eid].cap=0; edg[eid].flow=0; head[v]=eid++;
}

bool inq[MAXN];
bool spfa(int sNode,int eNode , int n){
    queue<int>q;
    for(int i=0; i<n; i++){
        inq[i]=false; cost[i]= -1;
    }
    cost[sNode]=0; inq[sNode]=1; pre[sNode]=-1;
    q.push(sNode);
    while(!q.empty()){
        int u=q.front(); q.pop();
        inq[u]=0;
        for(int i=head[u]; i!=-1; i=edg[i].next){
            int v=edg[i].to;
            if(edg[i].cap-edg[i].flow>0 && cost[v]<cost[u]+edg[i].cost){ //在满足可增流的情况下，最小花费
                cost[v] = cost[u]+edg[i].cost;
                pre[v]=i;   //记录路径上的边
                if(!inq[v])
                    q.push(v),inq[v]=1;
            }
        }
    }
    return cost[eNode]!=-1;    //推断有没有增广路
}
//反回的是最大流，最小花费为minCost
int minCost_maxFlow(int sNode,int eNode ,int& minCost , int n){
    int ans=0;
    while(spfa(sNode,eNode , n)){

        for(int i=pre[eNode]; i!=-1; i=pre[edg[i^1].to]){
            edg[i].flow+=1; edg[i^1].flow-=1;
            minCost+=edg[i].cost;
        }
        ans++;
        if(ans==2)break;
    }
    return ans;
}
int main(){
    int n,mapt[35][35];
    while(scanf("%d",&n)>0){
        init();
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            scanf("%d",&mapt[i][j]);
        int s = 0 , t = n*n-1;

        for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
        if(i||j){
            addEdg(i*n+j , i*n+j+n*n , 1 , mapt[i][j]);
            if(j+1<n) addEdg(i*n+j+n*n, i*n+j+1 , 1 , 0 );
            if(i+1<n) addEdg(i*n+j+n*n, (i+1)*n+j , 1 , 0);
        }
        else{
            addEdg(s , 1 , 1,0) , addEdg(s , n , 1 , 0);
        }
        int maxCost=mapt[0][0];
        if(n>1) maxCost+=mapt[n-1][n-1];
        minCost_maxFlow(s , t , maxCost , n*n*2);
        printf("%d
",maxCost);
    }
}