洛谷 P1833 樱花

题目传送门

解题思路:

就是完全背包和多重背包的混合.处理时间的时候注意一下就行了

AC代码:

#include<iostream>
#include<cstdio>

using namespace std;

string l,l1;
int h1,h2,m1,m2,t,n,w[200001],c[200001],p1,f[1001];
bool p,vis[200001];

inline void find_time() {
    cin >> l >> l1;
    for(int i = 0;i < l.length(); i++) {
        if(l[i] == ':') {
            p = 1;
            continue;
        }
        if(!p) h1 = h1 * 10 + (l[i] - '0');
        if(p) m1 = m1 * 10 + (l[i] - '0');
    }
    p = 0;
    for(int i = 0;i < l1.length(); i++) {
        if(l1[i] == ':') {
            p = 1;
            continue;
        }
        if(!p) h2 = h2 * 10 + (l1[i] - '0');
        if(p) m2 = m2 * 10 + (l1[i] - '0');
    }
    t = (h2 - h1) * 60 + (m2 - m1);
    m2 = 0;
}

int main() {
    find_time();
    scanf("%d",&n);
    for(int i = 1;i <= n; i++) {
        scanf("%d%d%d",&h1,&h2,&m1);
        if(m1 == 0) {
            w[++m2] = h1;
            c[m2] = h2;
            vis[m2] = 1;
        }
        else {
            for(int j = 1;j <= m1; j *= 2) {
                w[++m2] = h1 * j;
                c[m2] = h2 * j;
                m1 -= j;
            }
            if(m1 != 0) {
                w[++m2] = h1 * m1;
                c[m2] = h2 * m1;
            }
        }
    }
    for(int i = 1;i <= m2; i++) {
        if(!vis[i])
            for(int j = t;j >= w[i]; j--)
                f[j] = max(f[j],f[j-w[i]] + c[i]);
        else
            for(int j = w[i];j <= t; j++)
                f[j] = max(f[j],f[j-w[i]] + c[i]);
    }
    printf("%d",f[t]);
    return 0;
}