UVAlive 7061 Dire Wolf （区间DP）

题目链接

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 direwolf. 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.
Hint: In 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.
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

题意：

第一行一个T，表示测试组数；

第二行一个数n，表示有n匹狼；

第三行n个数，存在数组a[]里，表示杀死这匹狼你承受的基础伤害（还有附带伤害，下一行n个数表示，存在数组b[]里，杀死该匹狼还要承受的附带伤害为b[i-1]+b[i+1]，如果旁边的狼在之前被杀死了，那么附带伤害就是左右离他最近的狼的b[pos]之和）；

第四行n个数表示附带伤害，存在b[]数组里；

最后输出杀死所有的狼你能承受的最小伤害为多少。

分析：此题为区间dp，我们用dp[i][j]表示杀死i到j区间内所有狼所承受的最小伤害，对于一个k (i<=k<=j)表示这个区间内最后杀死这匹狼，那么dp转移方程就可以写成这样：dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]),i<=k<=j，由于k是最后杀死的狼，所以杀死这匹狼承受的附带伤害自然是第i-1匹狼和第j+1匹狼的附带伤害之和。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
#define inf 1e8
using namespace std;
ll dp[202][202],a[202],b[202];
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(j>=i) dp[i][j]=inf;
                else dp[i][j]=0;
            }
        }
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        b[0]=b[n+1]=0;
        for(int i=1;i<=n;i++) scanf("%lld",&b[i]);
        for(int i=1;i<=n;i++)
            dp[i][i]=a[i]+b[i-1]+b[i+1];
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j+i<=n;j++)
            {
                for(int k=j;k<=j+i;k++)
                    dp[j][j+i]=min(dp[j][j+i],dp[j][k-1]+dp[k+1][j+i]+a[k]+b[j-1]+b[j+i+1]);
            }
        }
        printf("Case #%d: %lld\n",ca,dp[1][n]);
    }
    return 0;
}