HDU 2686 Matrix (maximum cost Stream)

HDU 2686 Matrix (maximum cost Stream)
MatrixTime 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 shocould 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 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 before t the start and end.

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

Output For each test case output the maximal values yifenfei can get.
Sample Input

210 35 10310 3 32 5 36 7 1051 2 3 4 52 3 4 5 63 4 5 6 74 5 6 7 85 6 7 8 9

Sample Output

284680

Author yifenfei
Source ZJFC 2009-3 Programming Contest. From (0, 0) to (n-1, n-1) (left-to-right or top-to-bottom only) and then back to (0, 0) (right-to-left or bottom-up only) what is the maximum sum of vertices? Each vertex can only go once. Solution: Find the two greatest paths from (0, 0) to (n-1, n-1. Each edge capacity is 1, cost: For the vertex itself, the edge weight is the vertex weight, and the edge weight of non-vertex is 0.

# Include
 
  
# Include
  
   
# Include
   
    
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; // unit price} edg [MAXM]; int head [MAXN], eid; int pre [MAXN], cost [MAXN]; // point 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
    
     
Q; for (int I = 0; I
     
      
0 & cost [v]
      
        0) {init (); for (int I = 0; I
       
         1) maxCost + = mapt [n-1] [n-1]; minCost_maxFlow (s, t, maxCost, n * 2); printf ("% d \ n ", maxCost );}}