Good Bye 2014 F - New Year Shopping

F - New Year Shopping

对于一种特殊的不可逆的dp的拆分方法。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 4000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

int n, p, c[N], h[N], t[N], L[10001][N], dp[N], a[20001], b[20001], ans[20001];

vector<int> v[20001];
vector<int> qus[20001];

void solve(int T) {
    int down = max(T - p + 1, 1), up = min(20000, T + p - 1);
    memset(dp, 0, sizeof(dp));
    for(int k = T - 1; k >= down; k--) {
        for(int id : v[k]) {
            for(int j = N - 1; j >= c[id]; j--)
                dp[j] = max(dp[j - c[id]] + h[id], dp[j]);
        }
        memcpy(L[k - down], dp, sizeof(dp));
    }
    memset(dp, 0, sizeof(dp));
    for(int k = T; k <= up; k++) {
        for(int id : v[k]) {
            for(int j = N - 1; j >= c[id]; j--)
                dp[j] = max(dp[j - c[id]] + h[id], dp[j]);
        }
        for(int x : qus[k]) {
            for(int j = 0; j <= b[x]; j++)
                ans[x] = max(ans[x], dp[j] + L[k - T][b[x] - j]);
        }
    }
}

int main() {
    scanf("%d%d", &n, &p);
    for(int i = 1; i <= n; i++) {
        scanf("%d%d%d", &c[i], &h[i], &t[i]);
        v[t[i]].push_back(i);
    }
    int q; scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        scanf("%d%d", &a[i], &b[i]);
        qus[a[i]].push_back(i);
    }

    for(int i = 1; i <= 20000; i += p) {
        solve(i);
    }

    for(int i = 1; i <= q; i++) printf("%d\n", ans[i]);
    return 0;
}

/*
*/