Bzoj1855: [Scoi2010]股票交易
2018-02-21 09:55
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题面
BzojSol
设f[i][j]f[i][j]表示第ii天有jj张股票的最大收益转移很简单辣
# include <bits/stdc++.h> # define RG register # define IL inline # define Fill(a, b) memset(a, b, sizeof(a)) using namespace std; typedef long long ll; const int _(2010); IL int Input(){ RG int x = 0, z = 1; RG char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); return x * z; } int t, maxp, w, ap[_], bp[_], as[_], bs[_]; int f[_][_], ans = -2e9; int main(RG int argc, RG char* argv[]){ t = Input(), maxp = Input(), w = Input(); for(RG int i = 1; i <= t; ++i) ap[i] = Input(), bp[i] = Input(), as[i] = Input(), bs[i] = Input(); Fill(f, -127), f[0][0] = 0; for(RG int i = 1; i <= t; ++i){ for(RG int j = 0; j <= maxp; ++j) f[i][j] = f[i - 1][j]; for(RG int j = 0; j <= as[i] && j <= maxp; ++j) f[i][j] = max(f[i][j], -j * ap[i]); if(i <= w) continue; for(RG int j = 0; j <= maxp; ++j){ for(RG int k = j - 1; ~k && j - k <= as[i]; --k) f[i][j] = max(f[i][j], f[i - w - 1][k] - (j - k) * ap[i]); for(RG int k = j + 1; k <= maxp && k - j <= bs[i]; ++k) f[i][j] = max(f[i][j], f[i - w - 1][k] + (k - j) * bp[i]); } } for(RG int i = 0; i <= maxp; ++i) ans = max(ans, f[t][i]); printf("%d\n", ans); return 0; }
转移方程提出来与kk无关的然后单调队列辣
# include <bits/stdc++.h> # define RG register # define IL inline # define Fill(a, b) memset(a, b, sizeof(a)) using namespace std; typedef long long ll; const int _(2010); IL int Input(){ RG int x = 0, z = 1; RG char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); return x * z; } int t, maxp, w, ap[_], bp[_], as[_], bs[_]; int Q[_], head, tail, g[_]; int f[_][_], ans = -2e9; int main(RG int argc, RG char* argv[]){ t = Input(), maxp = Input(), w = Input(); for(RG int i = 1; i <= t; ++i) ap[i] = Input(), bp[i] = Input(), as[i] = Input(), bs[i] = Input(); Fill(f, -127), f[0][0] = 0; for(RG int i = 1; i <= t; ++i){ for(RG int j = 0; j <= maxp; ++j) f[i][j] = f[i - 1][j]; for(RG int j = 0; j <= as[i] && j & f389 lt;= maxp; ++j) f[i][j] = max(f[i][j], -j * ap[i]); if(i <= w) continue; head = 0, tail = -1; for(RG int j = 0; j <= maxp; ++j){ while(head <= tail && j - Q[head] > as[i]) ++head; if(head <= tail && j > Q[head]) f[i][j] = max(f[i][j], g[head] - j * ap[i]); while(head <= tail && g[tail] < f[i - w - 1][j] + j * ap[i]) --tail; Q[++tail] = j, g[tail] = f[i - w - 1][j] + j * ap[i]; } head = 0, tail = -1; for(RG int j = maxp; ~j; --j){ while(head <= tail && Q[head] - j > bs[i]) ++head; if(head <= tail && Q[head] > j) f[i][j] = max(f[i][j], g[head] - j * bp[i]); while(head <= tail && g[tail] < f[i - w - 1][j] + j * bp[i]) --tail; Q[++tail] = j, g[tail] = f[i - w - 1][j] + j * bp[i]; } } for(RG int i = 0; i <= maxp; ++i) ans = max(ans, f[t][i]); printf("%d\n", ans); return 0; }
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