Codevs 1043 Squares to take the number

$n the squares of the \times n$, each lattice has a natural number. One person in the square, can only go right or down. From the upper left to the lower right to walk two times, ask the maximum number of numbers can be obtained (each number is not taken, that is, two times through the same location can only get a value).

Consider walking together twice, each line passing through the segment is two segments assumed to be [a, b] and [C, D]. The current row state is [a, b] [C, d] when the transfer is from the previous line of B equals a of this line, and the previous line of D equals the C of this line.

In this case the complexity is $O (n^9) $, probably can not, but consider pruning, you can let $c \ge a$, because otherwise it can be two times the second half of the path as a $c \ge a$.

For example, 3-5 and 1-9, can be changed to 1-5 and 3-9.

This is probably $3\times 10^7$ at n = 10.

#include <bits/stdc++.h>using std::vector;const int maxn = 12;const int maxs = 1 << maxn;int dp[maxn][maxs];in  T a[maxn][maxn];int sum[maxn][maxn];struct status{int A, B, C, D;    int sumvalue (int i) {int res = sum[i][b + 1]-sum[i][a] + sum[i][d + 1]-sum[i][c];    if (c <= b) Res-= sum[i][b + 1]-sum[i][c];  return res;  }};vector<status> Status; #define REP (I, L, R) for (int i = l; i < R; i++) int statustable (int n) {int res = 0;    Rep (i0, 0, N) Rep (I1, I0, N) Rep (j0, I0, N) Rep (J1, J0, N) {status.push_back (status) {I0, I1, J0, J1});  res++; } return res;  int main () {int n;  scanf ("%d", &n);  int sCnt = statustable (n);  int x, y, V;    for (; scanf ("%d%d%d", &x, &y, &v), (x + y + V);)  A[x][y] = v;  for (int i = 1, i <= N; i++) {for (int j = 1; J <= N; j + +) Sum[i][j] = Sum[i][j-1] + a[i][j]; } Rep (i, 0, sCnt) if (status[i].a = = 0 && status[i].b >= status[i].c-1) {Dp[1][i] = Status[i].sumvalue (1); } for (int i = 2; I <= n; i++) {Rep (j, 0, SCnt) Rep (k, 0, SCnt) {if (status[j].b-status[k].a | | status[j].      D-STATUS[K].C) continue;    Dp[i][k] = Std::max (Dp[i][k], dp[i-1][j] + status[k].sumvalue (i));  }} int result = 0;  Rep (i, 0, sCnt) if (status[i].d = = N-1 && status[i].b = = n-1) {result = Std::max (dp[n][i], result);  } printf ("%d\n", result); return 0;}

　　

Codevs 1043 Squares to take the number