AtCoder Regular Contest 100 E - Or Plus Max
阿新 • • 發佈:2018-07-07
spa scanf cstring sin ans main swa const swap
一道很好的dp題
dp[K]存的是 i滿足二進制1屬於K二進制1位置 最大的兩個Ai
這樣dp[K]統計的兩個數肯定滿足(i | j) <= K
然後不斷做 update(dp[i | (1<<j)], dp[I])
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <ctime> using namespace std; typedef long long ll; const int N = 262200; const ll INF = 1e18; int A[N]; pair<int, int> dp[N]; void update(pair<int,int> &A, int x) { if(x > A.second) { A.second = x; if(A.second > A.first) { swap(A.second, A.first); } } } int main() { int n; while(~scanf("%d", &n)) { int nLen = (1<<n) - 1; for(int i = 0; i <= nLen; ++i) { scanf("%d", &A[i]); dp[i] = make_pair(A[i], -1); } for(int i = 0; i < n; ++i) { for(int j = 0; j <= nLen; ++j) { if( ((j >> i) & 1) == 0 ) { int newI = j | (1 << i); int t1 = dp[j].first; int t2 = dp[j].second; update(dp[newI], t1); update(dp[newI], t2); } } } // for(int i = 0; i <= nLen; ++i) printf("%d %d\n", dp[i].first, dp[i].second); int ans = -1; for(int i = 1; i <= nLen; ++i) { ans = max(ans, dp[i].first + dp[i].second); printf("%d\n", ans); } } return 0; }
AtCoder Regular Contest 100 E - Or Plus Max