Matrix

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

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

从左上到右下在走回左上，每个点只能经过一次，问最大的权值和是多少

可以当做成从左上到右下走过两次，至于每个点只能走过一次，需要拆点，每个点i到对应的i点容量是1，使得该点只能走过一次。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define maxn 3000
#define INF 0x3f3f3f3f
struct node
{
    int v , w , s ;
    int next ;
} p[3000000];
int a[50][50] , head[maxn] , cnt ;
int pre[maxn] , vis[maxn] , dis[maxn] ;
queue <int> q ;
void add(int u,int v,int w,int s)
{
    p[cnt].v = v ; p[cnt].w = w ; p[cnt].s = s ;
    p[cnt].next = head[u] ; head[u] = cnt++ ;
    p[cnt].v = u ; p[cnt].w = 0 ; p[cnt].s = -s ;
    p[cnt].next = head[v] ; head[v] = cnt++ ;
}
int spfa(int s,int t)
{
    int u , v , i ;
    memset(dis,INF,sizeof(dis));
    dis[s] = 0 ; vis[s] = 1 ;
    pre[s] = pre[t] = -1 ;
    while( !q.empty() )
        q.pop();
    q.push(s) ;
    while( !q.empty() )
    {
        u = q.front();
        q.pop();
        vis[u] = 0 ;
        for(i = head[u] ; i != -1 ; i = p[i].next)
        {
            v = p[i].v ;
            if( p[i].w && dis[v] > dis[u] + p[i].s )
            {
                dis[v] = dis[u] + p[i].s ;
                pre[v] = i ;
                if( !vis[v] )
                {
                    vis[v] = 1 ;
                    q.push(v) ;
                }
            }
        }
    }
    if( pre[t] == -1 )
        return 0 ;
    return 1 ;
}
void f(int s,int t)
{
    memset(pre,-1,sizeof(pre));
    memset(vis,0,sizeof(vis));
    int i , ans = 0 , min1 ;
    while( spfa(s,t) )
    {
        min1 = INF ;
        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])
            if( p[i].w < min1 )
                min1 = p[i].w ;
        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])
        {
            p[i].w -= min1 ;
            p[i^1].w += min1 ;
            ans += p[i].s ;
        }
    }
    printf("%d\n", -ans);
}
int main()
{
    int i , j , n , b , temp;
    while(scanf("%d", &n)!=EOF)
    {/*和D相同，不过这一个题换成了每个点只能经过一次，要求从左上走到右下，再走回左上，可以认为是从左上走到右下两次 */
        memset(head,-1,sizeof(head));
        cnt = 0 ; temp = n*n ;
        for(i = 1 ; i <= n ; i++)
            for(j = 1 ; j <= n ; j++)
            {
                scanf("%d", &a[i][j]);
                b = (i-1)*n + j ;
                if(b == 1 || b == temp)
                    add(b,b+temp,1,0);//源点和汇点会被走两次
                add(b,b+temp,1,-a[i][j]);//拆点操作，使得每个点只会被经过一次。
            }
        for(i = 1 ; i <= n ; i++)
            for(j = 1 ; j <= n ; j++)
            {
                b = (i-1)*n+j ;
                if(i > 1)
                    add(b-n+temp,b,1,0);
                if(j > 1)
                    add(b-1+temp,b,1,0);
            }
        add(0,1,2,0);
        add(2*n*n,2*n*n+1,2,0);
        f(0,2*n*n+1);
    }
    return 0;
}

hdu2686--Matrix(拆点+最大费用),布布扣,bubuko.com