掛飾 BZOJ - 4247

題目傳送門

思路：一個揹包問題，這個揹包問題和一般的揹包問題不同的是這個揹包的容量是可以變化的，所有我們定義DP[i][j]代表安裝i掛鉤以後還剩下j個掛鉤，然後遞推式就是dp[i][j] = max(dp[i - 1][j], dp[i - 1][max(j - p[i].a, 0) + 1] + p[i].b)

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define MAXN 4010
#define MAXE 5
#define INF 2100000000
#define MOD 100003
#define LL long long
#define ULL unsigned long long
#define pi 3.14159

using namespace std;

struct Node {
    int a;
    int b;
}p[MAXN];

bool cmp(const Node &p1, const Node &p2) {
    return p1.a > p2.a;
}

LL dp[MAXN][MAXN];

int main() {
    //std::ios::sync_with_stdio(false);
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", &p[i].a, &p[i].b);
    }
    sort(p + 1, p + n + 1, cmp);
    for (int i = 0; i <= n; ++i) {
        for (int j = 0; j <= n + 1; ++j) {
            dp[i][j] = -INF;
       }
    }
    dp[0][1] = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j <= n; ++j) {
            dp[i][j] = max(dp[i - 1][j], dp[i - 1][max(j - p[i].a, 0) + 1] + p[i].b);
        }
    }
    LL max_sum = 0;
    for (int i = 0; i <= n; ++i) {
        max_sum = max(max_sum, dp[n][i]);
    }
    printf("%lld\n", max_sum);
    return 0;
}