标签:lin class freopen ast 而且 ota row int pos
Dire Wolf
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 2578 Accepted Submission(s): 1509
Problem Description
Dire wolves, also known as Dark wolves, are 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
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 could 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
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 survive.
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.
Input
The 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.
Output
For 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
2
3
3 5 7
8 2 0
10
1 3 5 7 9 2 4 6 8 10
9 4 1 2 1 2 1 4 5 1
Sample Output
Case #1: 17
Case #2: 74
Hint
In the ?rst 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
有趣的一道区间dp,看到这种题也想到了是区间dp,一开始想表示dp[i][j]为消灭掉i-j之间的狼所受到的最小伤害,可后来发现一个问题,我们如何表示某个点的伤害能力呢,只要消灭的先后顺序不同受到的伤害就可能会发生改变,如何控制这个变量。我们可以在分割区间时把中间的分割点k看作是这个区间最后一个消灭的狼,[i,j]内k肯定要遍历一遍而且一定有一个是最后杀死的,所以可以这样表示,此时 dp[i][j]=dp[i][k-1]+dp[i][k+1]+a[k]+b[i-1]+b[j+1];
我们把k留在最后就不必在两个子区间内重复计算了,只需要额外加上k造成的伤害,k留在最后所以左边的就是i-1,右边的就是j+1了.
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define inf 0x3f3f3f3f
4 int a[205],b[205];
5 int dp[205][205];
6 int main()
7 {
8 int T,N,m,i,j,k;
9 #ifdef ONLINE_JUDGE
10 #else
11 freopen("in.txt","r",stdin);
12 #endif
13 cin>>T;
14 for(int cas=1;cas<=T;++cas)
15 {
16 scanf("%d",&N);
17 for(i=1;i<=N;++i) scanf("%d",&a[i]);
18 for(i=1;i<=N;++i) scanf("%d",&b[i]);
19 a[0]=b[0]=a[N+1]=b[N+1]=0;
20 memset(dp,0,sizeof(dp));
21 dp[0][0]=0;
22 for(i=1;i<=N;++i) dp[i][i]=a[i]+b[i-1]+b[i+1];
23 for(int len=2;len<=N;++len)
24 {
25 for(i=1;i+len-1<=N;++i)
26 {
27 j=i+len-1;
28 dp[i][j]=inf;
29 for(k=i;k<=j;++k)
30 {
31 dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]);
32 }
33 }
34 }
35 printf("Case #%d: %d\n",cas,dp[1][N]);
36 }
37 return 0;
38 }
HDU 5115 区间dp
标签:lin class freopen ast 而且 ota row int pos
原文地址:http://www.cnblogs.com/zzqc/p/7295834.html