HDU Max Sum plus plus---dp+ scrolling array

HDU 1024

The main topic: given m and N and N number, the maximum number of m consecutive sub-series of n is required, which requires the sub-sequence not to be handed.

Problem-Solving ideas:<1> dynamic programming, defining state Dp[i][j] The value of the I-satin sequence that represents the number of J before the sequence, where the sub-sequence of I includes A[j],

Max (Dp[m][k]), m<=k<=n is the result of the request

<2> initial state: dp[i][0] = 0, dp[0][j] = 0;

<3> State transfer:

Decision: A[j] to be a sub-segment, or to be followed by a previous sub-paragraph

Equation:

A[J] directly after the preceding sub-paragraph, then dp[i][j] = dp[i][j-1] + a[j];

A[j] himself as a sub-paragraph, first to obtain the largest number of K in front of the i-1 section and, plus a[j],

i.e. Dp[i][j] = max (Dp[i-1][k]) + a[j], where 1=<k<=j-1;

i.e. Dp[i][j] = max (Dp[i][j-1] + a[j], Max (Dp[i-1][k]) + a[j]), where k>=1&&k<=j-1;

<4> Space Complexity Optimization:

This is because n can reach 100 0000 and cannot be dp[n][n], so you need to use a scrolling array to optimize the space.

As you can see from the state transition equation, the update of the state Dp[i][j] with the current line position J only needs to take advantage of the first j-1 state maximum of the previous row

So you need up to two rows of arrays

<5> Time Complexity optimization:

Each time you update dp[i][j], you need to find the maximum value of the previous J-1 state of the previous line, so the complexity will reach O (n^3);

In fact, each time you update the current row's dp[i][j], the maximum value of its corresponding number of first j is saved in Pre[j],

This allows the next row to be updated directly with the pre[j] saved by the current row, instead of searching,

Therefore, time complexity can be optimized to O (n^2)

/*HDU Max Sum plus plus---dp+ scrolling array*/#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intMAXN =1000005;intDP[MAXN], PRE[MAXN];//Pre[j] = dp[i-1][intA[MAXN];intm, N;intMain () {#ifdef _local freopen ("D:\\input.txt","R", stdin);#endif     while(SCANF ("%d%d", &m, &n) = =2){        //input         for(inti =1; I <= N; ++i) {scanf ("%d", A +i); }//For (i)        intMAX; //Defines the precursor of the first state of the first row, at which I is equivalent to 1,dp[1][0]dp[0] =0; //initialize the pre array to be all 0, at which point the pre array should have a row of 0,memset (PRE,0,sizeofpre);  for(inti =1; I <= m; ++i) {            //The inner loop does is to update the state of the current line I, noting that J >= I makes senseMAX= -10e8;//initializes Max to a very small number             for(intj = i; J <= N; ++j) {                /*the PRE[J in the state transition equation] holds the maximum value of the number of first J of the previous line, i.e. pre[j] = max (Dp[i-1][k]), 1<=k<=j                    If A[J] directly belongs to paragraph I, then dp[i][j] = dp[i][j-1] + a[j] here is dp[j] = Dp[j-1]+a[j] If A[J] from a paragraph, you need to find the maximum value of dp[i-1][k], wherein 1<=k<=j-1, that is pre[j-1]*/Dp[j]= Max (Dp[j-1], Pre[j-1]) +A[j]; /*below the pre array and the slightly different above pre[j-1] save is the minimum value of the previous j-1 number of the previous line (line i-1) and the following pre [J-1] After the update is saved is the maximum number of previous j-1 of the current row, update is used for the next row of the DP update, that is, I will be +1 before the use of another note max is the last loop updated value, that is, the current line (i) of the former J 1-digit maximum value*/pre[j-1] = MAX;//update the Pre, save the maximum number of j-1 for the current row, and prepare for the DP update for the next line                if(Dp[j] >MAX) {                    //Max recorded here is the maximum value of the first J bit of the current row//to the next inner Loop J +1, which is why Max can be used to update pre[j-1]MAX = Dp[j];//Max records the maximum value of the front J bits of the current row                }            }        }//For (i)printf"%d\n", MAX); }    return 0;}

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HDU Max Sum plus plus---dp+ scrolling array