Max Sum plus plus---hdu1024 (dynamic programming for maximum sum of M segments)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1024

Test instructions is the number of n is divided into M-segment, the largest and;

DP[I][J] means that the J number is divided into the I segment, the result of selecting the number of J, rather than the current optimal solution,

So consider whether the number of numbers I have is a paragraph or a paragraph from the front.

So Dp[i][j]=max (Dp[i][j-1]+a[j], max+a[j]); Where Max divides the first j-1 number into the maximum value in the I-1 segment;

Because the problem of N is 100w,m do not know how much, so open a two-dimensional array may not be because only each state is related to its previous state, so we can use the DP "2" "N" to implement the scrolling array;

Xor ^ is the same as 0, the difference is 1, you can use ^1 to convert the state;

The problem of boundary should be considered;

#include <cstdio>#include<cstring>#include<iostream>#include<cmath>#include<vector>#include<algorithm>using namespacestd;#defineN 1000050#defineMOD 1000000007#defineMet (A, b) memset (A, B, sizeof (a))#defineINF 0x3f3f3f3ftypedefLong LongLL;intN, M, dp[2][n], a[n];intMain () { while(SCANF ("%d%d", &m, &n)! =EOF) {Met (DP,0); Met (A,0);  for(intI=1; i<=n; i++) scanf ("%d", &A[i]); intMax, ans, k; K=0;  for(intI=1; i<=m; i++) {k= k^1; Dp[k][i]= dp[k^1][i-1]+a[i];///Select the number of I, divided into paragraph I, so can only be a paragraph, then can only write;Max= dp[k^1][i-1]; Ans=Dp[k][i];  for(intj=i+1; j<=n; J + +)///J to start from I+1 because: to be divided into segments I must have at least I number, there is a j-1 below;{Max= Max (max, dp[k^1][j-1]);///Max divides the number of previous j-1 into the maximum value in the i-1 segment;Dp[k][j]= MAX (max + a[j], dp[k][j-1] + a[j]);///himself into a paragraph (Max); and the preceding paragraph (dp[i][j-1]);ans=Max (Dp[k][j], ans); }} printf ("%d\n", ans); }    return 0;}

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Max Sum plus plus---hdu1024 (dynamic programming for maximum sum of M segments)