思路讲解
注意石子是一个圈,这个时候化曲为直,化为长度为两倍的序列
然后区间dp一般是O(n**3)的,因为序列的合并可能发生在任何地方,不一定是头和尾
AC代码
AC https://www.luogu.com.cn/record/197959001
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| #include <iostream> #include <cstring> #include <algorithm> #include <deque> #include <queue> #include <vector> #include <set> #include <map> #include <unordered_map> #include <cmath> #include <bitset> #include <iterator> #include <random> #include <iomanip> #include <cctype> #include <array>
typedef long long ll; typedef std::pair<ll,ll> pll; typedef std::array<ll,3> arr; const ll MAXN=217;
ll N,T,A[MAXN],dp_min[MAXN][MAXN],dp_max[MAXN][MAXN],sumA[MAXN];
int main() { std::ios::sync_with_stdio(false); std::cin.tie(0);std::cout.tie(0); std::cin>>N; for(int i=1;i<=N;++i){ std::cin>>A[i]; sumA[i]=A[i]+sumA[i-1]; } for(int i=N+1;i<=2*N;++i){ A[i]=A[i-N]; sumA[i]=A[i]+sumA[i-1]; }
for(int i=1;i<=2*N;++i){ dp_min[i][i] = 0; dp_max[i][i] = 0; }
for(int len=2;len<=N;++len){ for(int i=1;i<=2*N-len+1;++i){ int j = i + len - 1; dp_min[i][j] = 1e18+7; dp_max[i][j] = 0; for(int k=i;k<j;++k){ ll cost = sumA[j] - sumA[i-1]; dp_min[i][j] = std::min(dp_min[i][j], dp_min[i][k] + dp_min[k+1][j] + cost); dp_max[i][j] = std::max(dp_max[i][j], dp_max[i][k] + dp_max[k+1][j] + cost); } } }
ll min_ans = 1e18+7; ll max_ans = 0; for(int i=1;i<=N;++i){ min_ans = std::min(min_ans, dp_min[i][i+N-1]); max_ans = std::max(max_ans, dp_max[i][i+N-1]); } std::cout<<min_ans<<"\n"; std::cout<<max_ans<<"\n"; return 0; }
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心路历程(WA,TLE,MLE……)
WA
https://www.luogu.com.cn/record/197954954
序列的合并可能发生在任何地方,不一定是头和尾
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| #include <iostream> #include <cstring> #include <algorithm> #include <deque> #include <queue> #include <vector> #include <set> #include <map> #include <unordered_map> #include <cmath> #include <bitset> #include <iterator> #include <random> #include <iomanip> #include <cctype> #include <array>
typedef long long ll; typedef std::pair<ll,ll> pll; typedef std::array<ll,3> arr; const ll MAXN=217;
ll N,T,A[MAXN],dp[MAXN][MAXN],sumA[MAXN];
int main() { std::ios::sync_with_stdio(false); std::cin.tie(0);std::cout.tie(0); std::cin>>N; for(int i=1;i<=N;++i){ std::cin>>A[i]; sumA[i]=A[i]+sumA[i-1]; } for(int i=N+1;i<=2*N;++i){ A[i]=A[i-N]; sumA[i]=A[i]+sumA[i-1]; } for(int i=2*N;i>=1;--i){ for(int j=i+1;j<=2*N;++j){ dp[i][j]=std::min(dp[i+1][j]+sumA[j]-sumA[i]+A[i], dp[i][j-1]+sumA[j-1]-sumA[i-1]+A[j]); } } ll ans=1e18+7; for(int i=1;i<=N;++i){ ans=std::min(dp[i][i+N-1],ans); } std::cout<<ans<<"\n"; std::memset(dp, 0, sizeof(dp)); for(int i=2*N;i>=1;--i){ for(int j=i+1;j<=2*N;++j){ dp[i][j]=std::max(dp[i+1][j]+sumA[j]-sumA[i]+A[i], dp[i][j-1]+sumA[j-1]-sumA[i-1]+A[j]); } } ans=0; for(int i=1;i<=N;++i){ ans=std::max(dp[i][i+N-1],ans); } std::cout<<ans<<"\n"; return 0; }
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