Hdu 5115 Dire Wolf (interval DP), hdu5115

Hdu 5115 Dire Wolf (interval DP), hdu5115
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, are extraordinarily large and powerful wolves. distinct, if not all, Dire Wolves appear to originate from Draenor.
Dire wolves look like normal wolves, but these creatures are of nearly twice the size. these powerful beasts, 8-9 feet long and weighing 600-800 pounds, are the most well-known orc mounts. as tall as a man, these great wolves have long tusked jaws that look like they cocould snap an iron bar. they have burning red eyes. dire wolves are mottled gray or black in color. dire wolves thrive in the northern regions of Kalimdor and in Mulgore.
Dire wolves are efficient pack hunters that kill anything they catch. They prefer to attack in packs, surrounding and flanking a foe when they can.
-Wowpedia, Your wiki guide to the World of Warcra Repository

Matt, an adventurer from the Eastern Kingdoms, meets a pack of dire wolves. there are N wolves standing in a row (numbered with 1 to N from left to right ). matt has to defeat all of them to keep ve.

Once Matt defeats a dire wolf, he will take some damage which is equal to the wolf's current attack. as gregarious beasts, each dire wolf I can increase its adjacent wolves 'attack by bi. thus, each dire wolf I's current attack consists of two parts, its basic attack ai and the extra attack provided by the current adjacent wolves. the increase of attack is temporary. once a wolf is defeated, its adjacent wolves will no longer get extra attack from it. however, these two wolves (if exist) will become adjacent to each other now.

For example, suppose there are 3 dire wolves standing in a row, whose basic attacks ai are (3, 5, 7), respectively. the extra attacks bi they can provide are (8, 2, 0 ). thus, the current attacks of them are (5, 13, 9 ). if Matt defeats the second wolf first, he will get 13 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 has 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), 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 is the case number (starting from 1), 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: 74HintIn the first sample, Matt defeats the dire wolves from left to right. He takes 5 + 5 + 7 = 17 points of damage which is the least damage he has to take. 

 

Question: There are n wolves, each of which has two attributes. One is an attack force and the other is an added value. Each time a wolf is killed, the damage value is
The sum of the attack value of this wolf and the value-added sum of the two wolves next to it is the minimum damage to all wolves ..

Idea: dp [I] [j] indicates the minimum damage to all wolves in the interval I and j.
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 );

Here, a [I] is the attacking power of wolf I, and B [I] is the added value of wolf I.

K enumerated in the state equation represents the number of the wolf killed in interval I and 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;}