洛谷1417 烹调方案 dp 贪心

#洛谷 1417 dp

传送门

挺有趣的一道dp题目，看上去接近于0/1背包，但是考虑到取每个点时间不同会对最后结果产生影响，因此需要进行预处理

对于物品x和物品y，当时间为p时，先加x后加y的收益为 a[x]-(p+c[x])*b[x]+a[y]-(p+c[x]+c[y])*by

而先加y再加x的收益为 a[y]-(p+c[y])*b[y]+a[x]-(p+c[y]+c[x])*bx

化简这两个式子，不难发现对于x和y，如果满足 c[x]*b[y]<c[y]*b[x] ，那么x 一定优于 y

由以上推论即可得解，对于题目中所给的物品，将其按照以上顺序排序，在进行0/1背包，即可得解

#include <cstdio>
#include <cstring>
#include <algorithm>

const int maxn = 100000 + 100;
struct data {
    long long ai, bi, ci;
};
data p[60];
long long dp[maxn];
int t, n;

bool cmp(data aa, data bb) {
    return (aa.ci * bb.bi < aa.bi * bb.ci);
}

int main () {
    scanf("%d %d", &t, &n);
    for (int i = 1; i <= n; i++) scanf("%lld", &p[i].ai);
    for (int i = 1; i <= n; i++) scanf("%lld", &p[i].bi);
    for (int i = 1; i <= n; i++) scanf("%lld", &p[i].ci);
    std :: sort(p + 1, p + n + 1, cmp);
    for (int i = 1; i <= n; i++)
        for (int j = t; j >= p[i].ci; j--) 
            dp[j] = std :: max(dp[j], dp[j - p[i].ci] + (p[i].ai - j * p[i].bi));
    long long ans = 0;
    for (int i = 1; i <= t; i++) ans = std :: max(ans, dp[i]);
    printf("%lld", ans);

    return 0;
}

当时做这题时想尝试多次贪心取最优值的办法，，然而，最后只得了30分，虽然尝试未成功，但是面对贪心题目时，这也不失为一种方法

附上乱搞代码

#include <cstdio>
#include <cstring>
#include <algorithm>


const int maxn = 50 + 10;
int T, n;
struct data {
    int ai;
    int bi;
    int ci;
};
data p[maxn];
int dp[100000 + 10];

bool cmp1(data aa, data bb) {
    return(aa.bi < bb.bi);
}
bool cmp2(data aa, data bb) {
    return (aa.ci < bb.ci);
}
bool cmp3(data aa, data bb) {
    return (aa.ai > bb.ai);
}

bool cmp4(data aa, data bb) {
    return (aa.bi * aa.ci < bb.bi * bb.ci);
}




int main () {
    scanf("%d %d", &T, &n);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i].ai);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i].bi);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i].ci);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = T; j >= p[i].ci; j--) {
            dp[j] = std :: max(dp[j],dp[j - p[i].ci] + (p[i].ai - j * p[i].bi));
        }
    }
    for (int i = 1; i <= T; i++) ans = std :: max(ans, dp[i]);
    std :: sort(p + 1, p + n + 1, cmp1);
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        for (int j = T; j >= p[i].ci; j--) {
            dp[j] = std :: max(dp[j],dp[j - p[i].ci] + (p[i].ai - j * p[i].bi));
        }
    }
    for (int i = 1; i <= T; i++) ans = std :: max(ans, dp[i]);
    std :: sort(p + 1, p + n + 1, cmp2);
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        for (int j = T; j >= p[i].ci; j--) {
            dp[j] = std :: max(dp[j],dp[j - p[i].ci] + (p[i].ai - j * p[i].bi));
        }
    }
    for (int i = 1; i <= T; i++) ans = std :: max(ans, dp[i]);
    std :: sort(p + 1, p + n + 1, cmp3);
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        for (int j = T; j >= p[i].ci; j--) {
            dp[j] = std :: max(dp[j],dp[j - p[i].ci] + (p[i].ai - j * p[i].bi));
        }
    }
    for (int i = 1; i <= T; i++) ans = std :: max(ans, dp[i]);
    std :: sort(p + 1, p + n + 1, cmp4);
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
        for (int j = T; j >= p[i].ci; j--) {
            dp[j] = std :: max(dp[j],dp[j - p[i].ci] + (p[i].ai - j * p[i].bi));
        }
    }
    for (int i = 1; i <= T; i++) ans = std :: max(ans, dp[i]);
    printf("%d", ans);
    return 0;
}