hdu 4600 Harvest Moon（模拟）

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题目链接：hdu 4600 Harvest Moon
代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1005;int W, H, N, D, M, S[20];ll V[maxn], A[maxn];void init () {    memset(S, 0, sizeof(S));    scanf("%d%d%d%d%d", &W, &H, &N, &D, &M);    int w = W % 3, h = H % 3;    S[9] = (W/3) * (H/3);    S[w*3] += (H/3);    S[h*3] += (W/3);    if (w * h == 1) {        S[5]++; S[3] -= 2; S[1] += 2;    } else if (w * h == 4) {        S[8]++; S[6] -= 2; S[4] += 2;    } else if (w * h == 2) {        S[7]++; S[6]--; S[3]--; S[2] += 2;    }}ll solve (int in, int ot, int n) {    int d = D / n, c[20];    memset(V, 0, sizeof(V));    V[0] = M;    for (int i = 0; i < d; i++) {        int k = V[i] / in, area = 0, p = 9;        memset(c, 0, sizeof(c));        while (p * ot > in) {            int t = min(S[p] - c[p], k);            k -= t, c[p] += t;            area += t * p;            V[i] -= t * in;            if (k == 0) break;            p--;        }        V[i+1] = V[i] + area * ot;    }    return V[d];}ll solve (int in, int ot, int n, int m) {    int d = D, c[10], p = 9;    memset(c, 0, sizeof(c));    memset(V, 0, sizeof(V));    memset(A, 0, sizeof(A));    V[0] = M;    for (int i = 0; i < d; i++) {        int k = V[i] / in, nw = 0;        while (i + n <= d) {            int ti = (d - i - n) / m + 1;            if (1LL * p * ot * ti <= in) break;            int t = min(S[p] - c[p], k);            k -= t, c[p] += t;            nw += 1LL * t * p;            V[i] -= 1LL * t * in;            if (k == 0) break;            p--;        }        if (i + m <= d) {            V[i+m] += A[i] * ot;            A[i+m] += A[i];        }        if (nw) {            V[i+n] += nw * ot;            A[i+n] += nw;        }        V[i+1] += V[i];    }    return V[d];}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        ll ans = M;        int q, p, n, m;        for (int i = 0; i < N; i++) {            scanf("%d%d%d%d", &q, &p, &n, &m);            if (m == 0)                ans = max(ans, solve(q, p, n));            else                ans = max(ans, solve(q, p, n, m));        }        printf("%lld\n", ans);    }    return 0;}
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