Dire WolfTime
limit:5000/5000 MS (java/others) Memory limit:512000/512000 K (java/others)
Total submission (s): 353 Accepted Submission (s): 210
Problem Descriptiondire Wolves, also known as Dark Wolves, is extraordinarily large and powerful wolves. Many, if not all, Dire wolves appear to originate from Draenor.
Dire wolves look like normal wolves, but these creatures is of nearly twice the size. These powerful beasts, 8-9 feet long and weighing 600-800 pounds, is the most well-known ORC mounts. As tall as a man, these great wolves has a long tusked jaws that look like they could snap an iron bar. They has burning red eyes. Dire wolves is mottled gray or black in color. Dire wolves thrive in the northern regions of Kalimdor and in Mulgore.
Dire Wolves is efficient pack hunters that kill anything they catch. They prefer to attack in packs, surrounding and flanking a foe when they can.
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Matt, an adventurer from the Eastern kingdoms, meets a pack of dire wolves. There is N wolves standing in a row (numbered with 1 to N from left to right). Matt has the defeat all of them to survive.
Once Matt defeats a dire wolf, he'll take some damage which are equal to the wolf's current attack. As gregarious beasts, each dire wolf I can increase their adjacent wolves ' attack by BI. Thus, each dire wolf I's current attack consists of the II parts, its basic attack AI and the extra attack provided by the CU Rrent adjacent wolves. The increase of attack is temporary. Once A wolf is defeated, its adjacent wolves would no longer get extra attack from it. However, these, wolves (if exist) would become adjacent to each and now.
For example, suppose there is 3 dire wolves standing in a row, whose basic attacks Ai is (3, 5, 7), respectively. The extra attacks bi they can provide is (8, 2, 0). Thus, the current attacks of them is (5, 13, 9). If Matt defeats the second Wolf first, he'll get points of damage and the Alive Wolves ' current attacks become (3, 15 ).
As an alert and resourceful adventurer, Matt can decide the order of the dire wolves he defeats. Therefore, he wants to know the least damage he had to take to defeat all the wolves.
Inputthe first line contains only one integer T, which indicates the number of test cases. For each test case, the first line contains only one integer N (2≤n≤200).
The second line contains N integers ai (0≤ai≤100000) and denoting the basic attack of each dire wolf.
The third line contains N integers bi (0≤bi≤50000), denoting the extra attack each dire wolf can provide.
Outputfor each test case, output a single line "Case #x: Y", where x was the case number (starting from 1), and Y is the least Damage Matt needs to take.
Sample Input
233 5 78 2 0101 3 5 7 9 2 4 6 8 109 4 1 2 1 2 1 4 5 1
Sample Output
Case #1:17Case #2: "Hintin the" RST sample, Matt defeats the dire wolves from left to right. He takes 5 + 5 + 7 = Points of damage which is the least damage he have to take.
Test instructions: There are n wolves, each wolf has two properties, a kind of attack, a value-added, each kill a wolf, the damage was
The wolf's attack value is the same as that of the two wolves next to it, and the minimum damage value for all wolves to be killed.
Idea: Dp[i][j] represents the smallest damage that all wolves in a range i,j have to kill.
The state transition equation is:
Dp[i][j]=min{dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]} (I<=K<J);
Where A[i] is the first Wolf's attack, B[i] for the first wolf's added value.
K, enumerated in the state equation, represents the number of the last wolf killed within the interval i,j.
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #define LL Long longusing namespace Std;const int inf=999999999;const int maxn=210;struct node{ int val,add;} A[maxn];int n,dp[maxn][maxn];void Initial () { memset (dp,-1,sizeof (DP));} void input () { scanf ("%d", &n); for (int i=1;i<=n;i++) scanf ("%d", &a[i].val); for (int i=1;i<=n;i++) scanf ("%d", &a[i].add); a[0].val=a[0].add=0; A[n+1].val=a[n+1].add=0;} int DP (int l,int r) { if (l>r) return 0; if (dp[l][r]!=-1) return dp[l][r]; int ans=inf; for (int i=l;i<=r;i++) ans=min (ANS,DP (l,i-1) +DP (i+1,r) +a[i].val+a[l-1].add+a[r+1].add); return Dp[l][r]=ans;} int main () { int T; scanf ("%d", &t); for (int co=1;co<=t;co++) { initial (); Input (); printf ("Case #%d:%d\n", CO,DP (1,n)); } return 0;}
Hdu 5115 Dire Wolf (interval dp)
