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题目大意：有n只狼，每只狼有一个自己攻击的属性，还有一个属性就是可以给左边和右边的狼提高攻击力。这个左边的意思是如果离得最近的那个死了，攻击力加给离得左边没死的最近的一个。
思路：一开始以为贪心可解，但是显然想简单了啊。后来知道了是区间dp，dp[i][j]代表在区间i到j内的最小伤害数。关键是划分区间，我们让设k为区间内最后死的那匹狼，那么区间内就有状态转移公式:dp[i][j] = min(dp[i][j], dp[i][k-1]+dp[k+1][j]+ num[k] + f[i-1]+ f[j+1)。因为我们假设k是区间内的最后一头狼，所以f[i-1]代表此时k所获得的加成的左边是来自i-1的， f[j+1]代表k此时所获得的加成的右边来自j+1的。通过记忆化我们就可以求出来最小的伤害是多少了。
Dire Wolf
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 342 Accepted Submission(s): 202
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.
Source
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
using namespace std;
inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}
const int maxn = 220;
int num[maxn];
int f[maxn];
int dp[maxn][maxn];
int dfs(int x, int y)
{
if(dp[x][y] != INF) return dp[x][y];
if(x > y) return 0;
for(int k = x; k <= y; k++)
{
dp[x][y] = min(dp[x][y], dfs(x, k-1)+dfs(k+1, y)+num[k]+f[x-1]+f[y+1]);
}
return dp[x][y];
}
int main()
{
int T;
T = read();
int Case = 1;
while(T--)
{
int n;
n = read();
f[0] = f[n+1] = 0;
for(int i = 1; i <= n; i++) num[i] = read();
for(int i = 1; i <= n; i++) f[i] = read();
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++) dp[i][j] = INF;
int sum = dfs(1, n);
printf("Case #%d: %d\n", Case++, sum);
}
return 0;
}
HDU 5115 Dire Wolf(记忆化搜索)
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原文地址：http://blog.csdn.net/xu12110501127/article/details/42873651
