Bupt 202 Chocolate Machine Dynamic planning

The 2011 Bupt Multi-School joint game H question

Bupt 202 Chocolate Mashine

The old OJ of the school to hang the agent to go on so here is a link to a hust on the topic:

Http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70168#problem/J

Are counted from 1 onwards

DP[I][J] Indicates the maximum value of money in the bottom-right corner of the chocolate line I, J Liecheg

Small cells with a value less than K "block"

if (I, J) itself is "blocking" then dp[i][j] = 0

if (i, j) both the upper and the left have "block" then dp[i][j] = A[i][j]

If only one of the above blocks is "block" then dp[i][j] = from the current grid to the left plus until you encounter a boundary or "block"

If only the left one block is "blocking" then dp[i][j] = starts with the current grid and adds to the top until it encounters a boundary or "block"

If both upper and left are not blocked then dp[i][j] = Dp[i-1][j] + dp[i][j-1]-dp[i-1][j-1] + a[i][j]

And then constantly update the maximum value in Dp[i][j]

It's not hard to follow my line of thought is more cumbersome WA 4 cannon tat

#include <cstdio>#include<cstdlib>#include<ctime>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<stack>#include<Set>#include<queue>#include<vector>using namespaceStd;typedefLong Longll;Const intMAXN =3010; ll DP[MAXN][MAXN];intA[MAXN][MAXN];intMain () {//freopen ("In.txt", "R", stdin);    intT; scanf ("%d", &T);  while(t--)    {        intN, M, K; scanf ("%d%d%d", &n, &m, &k);  for(inti =1; I <= N; i++)             for(intj =1; J <= M; J + +) scanf ("%d", &A[i][j]); ll ans=0;  for(inti =1; I <= N; i++)             for(intj =1; J <= M; J + +)            {                if(A[i][j] <k) Dp[i][j]=0; Else                {                    if(dp[i-1][J] = =0&& dp[i][j-1] ==0) Dp[i][j]=A[i][j]; Else if(I-1>=1&& dp[i-1][J] = =0) {Dp[i][j]=A[i][j]; intt =1;  while(J-t >=1&& DP[I][J-T]! =0) {Dp[i][j]+ = a[i][j-T]; T++; }                    }                    Else if(J-1>=1&& dp[i][j-1] ==0) {Dp[i][j]=A[i][j]; intt =1;  while(I-t >=1&& Dp[i-t][j]! =0) {Dp[i][j]+ = a[i-T]                            [j]; T++; }                    }                    Else{Dp[i][j]= dp[i-1][J] + dp[i][j-1]-dp[i-1][j-1] +A[i][j]; }                }                if(Dp[i][j] >ans) ans=Dp[i][j]; } printf ("%i64d\n", ans); }    return 0;}

Bupt 202 Chocolate Machine Dynamic planning