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  • Luogu1880 [NOI1995]石子合并 (区间DP)

    一个1A主席树的男人,沦落到褪水DP举步维艰

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
    #define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
    #define Max(a,b) ((a) > (b) ? (a) : (b))
    #define Min(a,b) ((a) < (b) ? (a) : (b))
    #define Fill(a,b) memset(a, b, sizeof(a))
    #define Abs(a) ((a) < 0 ? -(a) : (a))
    #define Swap(a,b) a^=b^=a^=b
    #define ll long long
    
    //#define ON_DEBUG
    
    #ifdef ON_DEBUG
    
    #define D_e_Line printf("
    
    ----------
    
    ")
    #define D_e(x)  cout << #x << " = " << x << endl
    #define Pause() system("pause")
    #define FileOpen() freopen("in.txt","r",stdin);
    
    #else
    
    #define D_e_Line ;
    #define D_e(x)  ;
    #define Pause() ;
    #define FileOpen() ;
    
    #endif
    
    struct ios{
        template<typename ATP>ios& operator >> (ATP &x){
            x = 0; int f = 1; char c;
            for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
            while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
            x*= f;
            return *this;
        }
    }io;
    using namespace std;
    
    const int N = 507;
    int f_max[N][N], f_min[N][N], a[N], sum[N];
    
    int main(){
    	FileOpen();
    	int n;
        io >> n;
        R(i,1,n){
        	io >> a[i];
        	a[i + n] = a[i];
        }
        R(i,1, n << 1) sum[i] = sum[i - 1] + a[i];
        
        R(len, 2, n){
            R(l, 1, n << 1){
                int r = l + len - 1;
                if(r > (n << 1)) continue;
                
                f_min[l][r] = 0x3f3f3f3f;
                
                R(k, l, r - 1){
                    f_min[l][r] = Min(f_min[l][r], f_min[l][k] + f_min[k+1][r] + sum[r] - sum[l-1]);
                    f_max[l][r] = Max(f_max[l][r], f_max[l][k] + f_max[k+1][r] + sum[r] - sum[l-1]);
                }
            }
        }
        
        int maxx = 0, minn = 0x7fffffff;
        R(i,1,n){
        	maxx = Max(maxx, f_max[i][i + n - 1]);
        	minn = Min(minn, f_min[i][i + n - 1]);
        }
        
        printf("%d
    %d", minn, maxx);
    
        return 0;
    }
    

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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11228243.html
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