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; }
时间: 2024-10-11 06:16:24