Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5115
Surface:
Dire WolfTime
limit:5000/5000 MS (java/others) Memory limit:512000/512000 K (java/others)
Total submission (s): 687 Accepted Submission (s): 391
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.
SOURCE2014ACM/ICPC Asia Beijing Station
Test instructions: To kill all wolves suffered the smallest damage, the wolf itself is fixed, the accumulation of good. Each time a wolf is killed, it will be attacked by the wolves remaining next to him. Similar to matrix multiplication, the interval length from small to large DP, the length of 2, as if cleverly avoided the special award. The original interval DP is also not very difficult, mainly to be able to build out this model.
Code:
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <map># Include <cmath> #include <cstdlib> #include <set> #include <algorithm> #include <string># Include <iomanip> #define LL long longusing namespace Std;int store[222];int dp[220][220];int main () {int T,n,ans, TMP,LIM;SCANF ("%d", &t), for (int. i=1;i<=t;i++) {printf ("Case #%d:", i); ans=0; scanf ("%d", &n); store[0]=store[n+1]=0; for (int j=0;j<n;j++) {scanf ("%d", &tmp); ans+=tmp; } for (int j=1;j<=n;j++) {scanf ("%d", &store[j]); } for (int j=1;j<=n;j++) dp[j][j]= (store[j-1]+store[j+1]); for (int j=2;j<=n;j++) {lim=n-j+1; for (int k=1;k<=lim;k++) {dp[k][k+j-1]=dp[k+1][k+j-1]+store[k-1]+store[k+j]; for (int m=k;m<=k+j-1;m++) {dp[k][k+j-1]=min (dp[k][k+j-1],dp[k][m-1]+dp[m+1][k+j-1]+store[k-1]+store[k+j]); }}} Ans+=dp[1][n]; printf ("%d\n", ans);} return 0;}
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HDU 5115Dire Wolf (interval dp)
