HDU 3401 Trade【单调队列+dp】
2012-08-27 16:49
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题意:已知股票每天的买入和卖出价格、买入上限和卖出上限以及最多能持有的股票数,问n天后的最大收益是多少。
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; const int oo = 1 << 30; const int maxn = 2010; int tcase; int t, maxp, w; int ap[maxn], bp[maxn]; int as[maxn], bs[maxn]; int dp[maxn][maxn]; int head, tail; int q[maxn][2]; void DP() { for (int i = 0; i < maxn; ++i) { for (int j = 0; j < maxn; ++j) { dp[i][j] = -oo; } } // 前w+1天只能买入 for (int i = 1; i <= w + 1; ++i) { for (int j = 0; j <= min(maxp, as[i]); ++j) { dp[i][j] = -ap[i] * j; } } for (int i = 2; i <= t; ++i) { /*for (int j = 0; j <= maxp; ++j) { dp[i][j] = max(dp[i][j], dp[i-1][j]); } if (i <= w + 1) continue; */ head = tail = 0; for (int j = 0; j <= maxp; ++j) { dp[i][j] = max(dp[i][j], dp[i-1][j]); if (i <= w + 1) continue; while (head < tail && q[tail-1][1] <= dp[i-w-1][j] + ap[i] * j) { tail--; } q[tail][0] = j; q[tail++][1] = dp[i-w-1][j] + ap[i] * j; while (head < tail && j - q[head][0] > as[i]) { head++; } dp[i][j] = max(dp[i][j], q[head][1] - ap[i] * j); } head = tail = 0; for (int j = maxp; j >= 0; --j) { if (i <= w + 1) continue; while (head < tail && q[tail-1][1] <= dp[i-w-1][j] + bp[i] * j) { tail--; } q[tail][0] = j; q[tail++][1] = dp[i-w-1][j] + bp[i] * j; while (head < tail && q[head][0] - j > bs[i]) { head++; } dp[i][j] = max(dp[i][j], q[head][1] - bp[i] * j); } } int ans = -oo; for (int i = 0; i <= maxp; ++i) { ans = max(ans, dp[t][i]); } printf("%d\n", ans); } int main() { scanf("%d", &tcase); while (tcase--) { scanf("%d%d%d", &t, &maxp, &w); for (int i = 1; i <= t; ++i) { scanf("%d%d%d%d", &ap[i], &bp[i], &as[i], &bs[i]); } DP(); } return 0; }
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