The problem is the interval DP.
The problem is that AI means that each wolf has its own attack power, BI indicates that it can provide the attack to the neighboring wolves, and in what order to kill the Wolf, the wolf can attack the person least.
And if the wolf is killed, it will not be able to provide damage to the neighboring wolves.
The dynamic transfer equation is dp[i][j]=min (dp[i][k]+dp[k][j]+a[k]+a[i]+a[j],dp[i][j]) (j<k<i);
I indicates that the first wolf is the right end (not yet chopped), J indicates that the head of the first wolf is the left end (not yet killed).
DP[I][J] records the smallest attack that I-j has chosen when I,j is not selected.
(This problem does me ....) Spent almost a day, in the middle of thinking about a daze, just this state. Still lazy,,,, but now do not think too clear ... )。
First, a, b array at both ends to add a bit, a[0]=0;b[0]=0;a[p+1]=0;b[p+1]=0;
Because to kill all the wolves, because every time there are two ends, so to kill it must be 0-p+1 within the number 1 to P Wolf, 1 left, p right end must have an endpoint. The final answer is dp[p+1][0];
Get back to the code before you finish the lesson:
1#include <stdio.h>2#include <algorithm>3#include <iostream>4#include <stdlib.h>5#include <string.h>6 using namespacestd;7 Long Longn=1e16;8typedefLong Longll;9 intMainvoid)Ten { One inti,j,k,p,q; All dp[ -][ -]; -ll a[ -]; -ll b[ -]; thescanf"%d",&k); - for(i=1; i<=k; i++) - { -scanf"%d",&p); +Memset (A,0,sizeof(a)); -memset (b,0,sizeof(b)); + for(j=1; j<=p; J + +) Ascanf"%lld",&a[j]); at for(j=1; j<=p; J + +) -scanf"%lld",&b[j]); - intn,m,t; -Memset (DP,0,sizeof(DP)); - for(n=2; n<=p+1; n++) - { in for(m=n-2; m>=0; m--) - { todp[n][m]=dp[n][n-1]+dp[n-1][m]+a[n-1]+b[n]+B[m]; + for(t=n-1; t>m; t--) - { the *Dp[n][m]=min (dp[n][t]+dp[t][m]+b[n]+a[t]+b[m],dp[n][m]); $ }Panax Notoginseng } - the } +printf"Case #%d:%lld\n", i,dp[p+1][0]); A } the return 0; + -}
Hdu5115-dire Wolf (interval dp)