hdu5115 区间DP

Posted 2020-08-06 十目

题意：有n只狼，每只狼有两种属性，一种攻击力一种附加值，每杀一只狼

受到的伤害值为这只狼的攻击值与它旁边的两只狼的附加值的和，求把所有狼都杀光受到的最小的伤害值。

题解：还是老问题，，，区间DP想到了但是担心枚举i~j区间中元素时，处理dp[i][k-1]的时候要顾及i-1位置的狼，其实根本不用，初始化的时候已经顾及到了就说明全部都顾及到了，，，

真是菜

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define PF(x) cout << "debug: " << x << " ";
#define EL cout << endl;
#define PC(x) puts(x);
typedef long long ll;
#define CLR(x, v) sizeof (x, v, sizeof(x))
using namespace std;
const int INF = 0x5f5f5f5f;
const int  N= 2e5 + 10;
const int mod=1e9 + 7;
const int maxn = 210;
using namespace std;
int T,n;
int dp[maxn][maxn],a[maxn],b[maxn];
int main() {
  // freopen("in.txt","r",stdin);
    cin>>T;
    int cas = 0;
   while(T--){
        cas++;
        scanf("%d",&n);
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        for(int i = 1;i <= n;i++)
            scanf("%d",&b[i]);
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= n;j++)
                dp[i][j] = 40000000+10;
        for(int i = 1;i <= n;i++)
            dp[i][i] = a[i] + b[i-1] + b[i+1];
        //dp[0][0] = b[1],dp[n+1][n+1] = b[n];
        for(int i = n - 1;i >= 1;i--)
            for(int j = i + 1;j <= n;j++)
                for(int k = i ;k <= j;k++){
                    if(k == i)
                        dp[i][j] = min(dp[i][j],dp[i+1][j]+a[i]+b[i-1]+b[j+1]);
                    else if(k == j)
                        dp[i][j] = min(dp[i][j],dp[i][j-1]+a[j]+b[i-1]+b[j+1]);
                    else
                    dp[i][j] = min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]);
                }
        printf("Case #%d: %d\n",cas,dp[1][n]);
   }
    return 0;
}