Interval DP Primer

First we need to know how the interval is done with DP, so let's look at the template.

1  for(inti =1; I <= N; i++) {//number of enumeration intervals2      for(intj =1; J <= can enumerate to the largest POS; J + +){3         intp = i + J-1;//represents the coordinates of the maximum value that can be reached at the moment4         if(P > N) Break;5          for(intK = J; K <= P; k++){6Dp[j][p] = min or Max (dp[j][p], dp[j][k] + dp[k +][p] +J to P Val);7         }8     }9}

View Code

① Stone Problem

http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=737

This is a very obvious template problem.

You can use sum[] in advance to maintain intervals, or you can maintain them in a tree-like array.

#include <cstdio>#include<algorithm>#include<cstring>using namespacestd;Const intMAXN = $+5;Const intINF =0x3f3f3f3f;intDP[MAXN][MAXN];intA[MAXN];intSUM[MAXN];intN;intMain () { while(SCANF ("%d", &n) = =1){         for(inti =1; I <= N; i++) scanf ("%d", A +i); Memset (DP,0,sizeof(DP)); memset (SUM,0,sizeof(sum));  for(inti =1; I <= N; i++) {Sum[i]= Sum[i-1] +A[i]; }         for(inti =2; I <= N; i++){             for(intj =1; J <= N-i +1; J + +){                intp = j + I-1; if(P > N) Break; DP[J][P]=inf;  for(intK = J; K <= P; k++) {Dp[j][p]= Min (Dp[j][p], dp[j][k] + dp[k +1][P] + sum[p]-sum[j-1]); }}} printf ("%d\n", dp[1][n]); }    return 0;}

View Code

Ii

Interval DP Primer